Maximize the value of $sqrt{x-x^2}+sqrt{cx-x^2}$ without using calculus












3












$begingroup$


Assume that $c$ is positive. How can we maximize the value of $sqrt{x-x^2}+sqrt{cx-x^2}$ with respect to $x$ without the use of calculus?



With calculus, we can easily find out that the max of the expression is when $x=frac{c}{c+1}$.



My attempt to the question is consider the expression as the distance between points. Below is the figure.
triangle
The question becomes finding the longest length of the red line. However, I have no idea how to proceed.






optimization triangle radicals substitution a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Assume that $c$ is positive. How can we maximize the value of $sqrt{x-x^2}+sqrt{cx-x^2}$ with respect to $x$ without the use of calculus?



    With calculus, we can easily find out that the max of the expression is when $x=frac{c}{c+1}$.



    My attempt to the question is consider the expression as the distance between points. Below is the figure.
    triangle
    The question becomes finding the longest length of the red line. However, I have no idea how to proceed.






    optimization triangle radicals substitution a.m.-g.m.-inequality







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Assume that $c$ is positive. How can we maximize the value of $sqrt{x-x^2}+sqrt{cx-x^2}$ with respect to $x$ without the use of calculus?



      With calculus, we can easily find out that the max of the expression is when $x=frac{c}{c+1}$.



      My attempt to the question is consider the expression as the distance between points. Below is the figure.
      triangle
      The question becomes finding the longest length of the red line. However, I have no idea how to proceed.






      optimization triangle radicals substitution a.m.-g.m.-inequality







      share|cite|improve this question











      $endgroup$




      Assume that $c$ is positive. How can we maximize the value of $sqrt{x-x^2}+sqrt{cx-x^2}$ with respect to $x$ without the use of calculus?



      With calculus, we can easily find out that the max of the expression is when $x=frac{c}{c+1}$.



      My attempt to the question is consider the expression as the distance between points. Below is the figure.
      triangle
      The question becomes finding the longest length of the red line. However, I have no idea how to proceed.






      optimization triangle radicals substitution a.m.-g.m.-inequality




      optimization triangle radicals substitution a.m.-g.m.-inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Oct 22 '18 at 21:41









      Michael Rozenberg

      99.3k1590189




      99.3k1590189










      asked Oct 14 '18 at 9:09









      WssWss

      534




      534






















          3 Answers
          3






          active

          oldest

          votes


















          0












          $begingroup$

          Using the Cauchy-Schwartz inequality



          $$
          left(sqrt{x-x^2}+sqrt{c x-x^2}right)^2le left(x+1-xright)left(x+c-xright) = c
          $$



          then



          $$
          sqrt{x-x^2}+sqrt{c x-x^2}lesqrt{c}
          $$






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            Let $x=frac{c}{c+1}t.$



            Thus, by AM-GM we obtain:
            $$sqrt{x-x^2}+sqrt{cx-x^2}=frac{sqrt{c}}{c+1}sqrt{t(c+1-ct)}+frac{sqrt{c}}{c+1}sqrt{ct(c+1-t)}leq$$
            $$leqfrac{sqrt{c}}{c+1}left(frac{t+c+1-ct}{2}+frac{ct+c+1-t}{2}right)=sqrt{c}.$$
            THe equality occurs for $t=1$, which says that we got a maximal value.






            share|cite|improve this answer









            $endgroup$









            • 2




              $begingroup$
              Hi Michael thanks again for you kind support during the suspension time. If you are interested I've raised a related discussion HERE. Cheers
              $endgroup$
              – gimusi
              Oct 14 '18 at 10:26






            • 1




              $begingroup$
              Hi @gimusi . I really don't like discussions in Meta. In my opinion this a place, where a limited number of users, which are destroying this site, trying to get a legitimation to do it.
              $endgroup$
              – Michael Rozenberg
              Oct 14 '18 at 12:38










            • $begingroup$
              I think that open discussion are the only way to find a compromise between different point of views. Thanks again. Bye
              $endgroup$
              – gimusi
              Oct 14 '18 at 12:46










            • $begingroup$
              Nice approach but somewhat cheating to take $x=frac{c}{c+1}t$ at the beginning.
              $endgroup$
              – Wss
              Oct 15 '18 at 4:44










            • $begingroup$
              @Wss Because we know that the equality occurs for $x=frac{c}{c +1}$ and it would be much more better if it will happen for some variable is equal to $1$.
              $endgroup$
              – Michael Rozenberg
              Oct 15 '18 at 4:47



















            1












            $begingroup$

            Assuming $$sqrt{x-x^2}+sqrt{cx-x^2}le sqrt{c}$$
            Squaring we get



            $$2sqrt{x-x^2}sqrt{cx-x^2}le 2x^2-x-cx+c$$
            squaring again and factorizing we get $$(cx-c+x)^2geq 0$$ which is true.
            Remark: We can only square the inequality if $$2x^2-x(c+1)+cgeq 0$$ this is true if $$0<cle 1$$ for $$0le xle c$$ or $$c>1$$ and $$0le xle 1$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @Sonnhard I think the right side of your second inequality can be negative. At least, you need to prove that it's not negative.
              $endgroup$
              – Michael Rozenberg
              Oct 14 '18 at 9:27












            • $begingroup$
              This is true, the condition is complicated.
              $endgroup$
              – Dr. Sonnhard Graubner
              Oct 14 '18 at 9:38










            • $begingroup$
              @Sonnhard But $c^2-6c+1$ can be positive, which says that the right side can be negative. No?
              $endgroup$
              – Michael Rozenberg
              Oct 14 '18 at 9:41










            • $begingroup$
              Yes with the inequalities $$x-x^2geq 0$$ and$$ cx-x^2geq 0$$ it simplifies to the given above
              $endgroup$
              – Dr. Sonnhard Graubner
              Oct 14 '18 at 9:53












            • $begingroup$
              @Sonnhard But how did you got it? By the way, there is a very nice proof that $2x^2-(c+1)x+1>0$. Try to find it.
              $endgroup$
              – Michael Rozenberg
              Oct 14 '18 at 9:59











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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            Using the Cauchy-Schwartz inequality



            $$
            left(sqrt{x-x^2}+sqrt{c x-x^2}right)^2le left(x+1-xright)left(x+c-xright) = c
            $$



            then



            $$
            sqrt{x-x^2}+sqrt{c x-x^2}lesqrt{c}
            $$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Using the Cauchy-Schwartz inequality



              $$
              left(sqrt{x-x^2}+sqrt{c x-x^2}right)^2le left(x+1-xright)left(x+c-xright) = c
              $$



              then



              $$
              sqrt{x-x^2}+sqrt{c x-x^2}lesqrt{c}
              $$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Using the Cauchy-Schwartz inequality



                $$
                left(sqrt{x-x^2}+sqrt{c x-x^2}right)^2le left(x+1-xright)left(x+c-xright) = c
                $$



                then



                $$
                sqrt{x-x^2}+sqrt{c x-x^2}lesqrt{c}
                $$






                share|cite|improve this answer









                $endgroup$



                Using the Cauchy-Schwartz inequality



                $$
                left(sqrt{x-x^2}+sqrt{c x-x^2}right)^2le left(x+1-xright)left(x+c-xright) = c
                $$



                then



                $$
                sqrt{x-x^2}+sqrt{c x-x^2}lesqrt{c}
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 5 '18 at 20:46









                CesareoCesareo

                8,6793516




                8,6793516























                    3












                    $begingroup$

                    Let $x=frac{c}{c+1}t.$



                    Thus, by AM-GM we obtain:
                    $$sqrt{x-x^2}+sqrt{cx-x^2}=frac{sqrt{c}}{c+1}sqrt{t(c+1-ct)}+frac{sqrt{c}}{c+1}sqrt{ct(c+1-t)}leq$$
                    $$leqfrac{sqrt{c}}{c+1}left(frac{t+c+1-ct}{2}+frac{ct+c+1-t}{2}right)=sqrt{c}.$$
                    THe equality occurs for $t=1$, which says that we got a maximal value.






                    share|cite|improve this answer









                    $endgroup$









                    • 2




                      $begingroup$
                      Hi Michael thanks again for you kind support during the suspension time. If you are interested I've raised a related discussion HERE. Cheers
                      $endgroup$
                      – gimusi
                      Oct 14 '18 at 10:26






                    • 1




                      $begingroup$
                      Hi @gimusi . I really don't like discussions in Meta. In my opinion this a place, where a limited number of users, which are destroying this site, trying to get a legitimation to do it.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 12:38










                    • $begingroup$
                      I think that open discussion are the only way to find a compromise between different point of views. Thanks again. Bye
                      $endgroup$
                      – gimusi
                      Oct 14 '18 at 12:46










                    • $begingroup$
                      Nice approach but somewhat cheating to take $x=frac{c}{c+1}t$ at the beginning.
                      $endgroup$
                      – Wss
                      Oct 15 '18 at 4:44










                    • $begingroup$
                      @Wss Because we know that the equality occurs for $x=frac{c}{c +1}$ and it would be much more better if it will happen for some variable is equal to $1$.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 15 '18 at 4:47
















                    3












                    $begingroup$

                    Let $x=frac{c}{c+1}t.$



                    Thus, by AM-GM we obtain:
                    $$sqrt{x-x^2}+sqrt{cx-x^2}=frac{sqrt{c}}{c+1}sqrt{t(c+1-ct)}+frac{sqrt{c}}{c+1}sqrt{ct(c+1-t)}leq$$
                    $$leqfrac{sqrt{c}}{c+1}left(frac{t+c+1-ct}{2}+frac{ct+c+1-t}{2}right)=sqrt{c}.$$
                    THe equality occurs for $t=1$, which says that we got a maximal value.






                    share|cite|improve this answer









                    $endgroup$









                    • 2




                      $begingroup$
                      Hi Michael thanks again for you kind support during the suspension time. If you are interested I've raised a related discussion HERE. Cheers
                      $endgroup$
                      – gimusi
                      Oct 14 '18 at 10:26






                    • 1




                      $begingroup$
                      Hi @gimusi . I really don't like discussions in Meta. In my opinion this a place, where a limited number of users, which are destroying this site, trying to get a legitimation to do it.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 12:38










                    • $begingroup$
                      I think that open discussion are the only way to find a compromise between different point of views. Thanks again. Bye
                      $endgroup$
                      – gimusi
                      Oct 14 '18 at 12:46










                    • $begingroup$
                      Nice approach but somewhat cheating to take $x=frac{c}{c+1}t$ at the beginning.
                      $endgroup$
                      – Wss
                      Oct 15 '18 at 4:44










                    • $begingroup$
                      @Wss Because we know that the equality occurs for $x=frac{c}{c +1}$ and it would be much more better if it will happen for some variable is equal to $1$.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 15 '18 at 4:47














                    3












                    3








                    3





                    $begingroup$

                    Let $x=frac{c}{c+1}t.$



                    Thus, by AM-GM we obtain:
                    $$sqrt{x-x^2}+sqrt{cx-x^2}=frac{sqrt{c}}{c+1}sqrt{t(c+1-ct)}+frac{sqrt{c}}{c+1}sqrt{ct(c+1-t)}leq$$
                    $$leqfrac{sqrt{c}}{c+1}left(frac{t+c+1-ct}{2}+frac{ct+c+1-t}{2}right)=sqrt{c}.$$
                    THe equality occurs for $t=1$, which says that we got a maximal value.






                    share|cite|improve this answer









                    $endgroup$



                    Let $x=frac{c}{c+1}t.$



                    Thus, by AM-GM we obtain:
                    $$sqrt{x-x^2}+sqrt{cx-x^2}=frac{sqrt{c}}{c+1}sqrt{t(c+1-ct)}+frac{sqrt{c}}{c+1}sqrt{ct(c+1-t)}leq$$
                    $$leqfrac{sqrt{c}}{c+1}left(frac{t+c+1-ct}{2}+frac{ct+c+1-t}{2}right)=sqrt{c}.$$
                    THe equality occurs for $t=1$, which says that we got a maximal value.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Oct 14 '18 at 9:24









                    Michael RozenbergMichael Rozenberg

                    99.3k1590189




                    99.3k1590189








                    • 2




                      $begingroup$
                      Hi Michael thanks again for you kind support during the suspension time. If you are interested I've raised a related discussion HERE. Cheers
                      $endgroup$
                      – gimusi
                      Oct 14 '18 at 10:26






                    • 1




                      $begingroup$
                      Hi @gimusi . I really don't like discussions in Meta. In my opinion this a place, where a limited number of users, which are destroying this site, trying to get a legitimation to do it.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 12:38










                    • $begingroup$
                      I think that open discussion are the only way to find a compromise between different point of views. Thanks again. Bye
                      $endgroup$
                      – gimusi
                      Oct 14 '18 at 12:46










                    • $begingroup$
                      Nice approach but somewhat cheating to take $x=frac{c}{c+1}t$ at the beginning.
                      $endgroup$
                      – Wss
                      Oct 15 '18 at 4:44










                    • $begingroup$
                      @Wss Because we know that the equality occurs for $x=frac{c}{c +1}$ and it would be much more better if it will happen for some variable is equal to $1$.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 15 '18 at 4:47














                    • 2




                      $begingroup$
                      Hi Michael thanks again for you kind support during the suspension time. If you are interested I've raised a related discussion HERE. Cheers
                      $endgroup$
                      – gimusi
                      Oct 14 '18 at 10:26






                    • 1




                      $begingroup$
                      Hi @gimusi . I really don't like discussions in Meta. In my opinion this a place, where a limited number of users, which are destroying this site, trying to get a legitimation to do it.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 12:38










                    • $begingroup$
                      I think that open discussion are the only way to find a compromise between different point of views. Thanks again. Bye
                      $endgroup$
                      – gimusi
                      Oct 14 '18 at 12:46










                    • $begingroup$
                      Nice approach but somewhat cheating to take $x=frac{c}{c+1}t$ at the beginning.
                      $endgroup$
                      – Wss
                      Oct 15 '18 at 4:44










                    • $begingroup$
                      @Wss Because we know that the equality occurs for $x=frac{c}{c +1}$ and it would be much more better if it will happen for some variable is equal to $1$.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 15 '18 at 4:47








                    2




                    2




                    $begingroup$
                    Hi Michael thanks again for you kind support during the suspension time. If you are interested I've raised a related discussion HERE. Cheers
                    $endgroup$
                    – gimusi
                    Oct 14 '18 at 10:26




                    $begingroup$
                    Hi Michael thanks again for you kind support during the suspension time. If you are interested I've raised a related discussion HERE. Cheers
                    $endgroup$
                    – gimusi
                    Oct 14 '18 at 10:26




                    1




                    1




                    $begingroup$
                    Hi @gimusi . I really don't like discussions in Meta. In my opinion this a place, where a limited number of users, which are destroying this site, trying to get a legitimation to do it.
                    $endgroup$
                    – Michael Rozenberg
                    Oct 14 '18 at 12:38




                    $begingroup$
                    Hi @gimusi . I really don't like discussions in Meta. In my opinion this a place, where a limited number of users, which are destroying this site, trying to get a legitimation to do it.
                    $endgroup$
                    – Michael Rozenberg
                    Oct 14 '18 at 12:38












                    $begingroup$
                    I think that open discussion are the only way to find a compromise between different point of views. Thanks again. Bye
                    $endgroup$
                    – gimusi
                    Oct 14 '18 at 12:46




                    $begingroup$
                    I think that open discussion are the only way to find a compromise between different point of views. Thanks again. Bye
                    $endgroup$
                    – gimusi
                    Oct 14 '18 at 12:46












                    $begingroup$
                    Nice approach but somewhat cheating to take $x=frac{c}{c+1}t$ at the beginning.
                    $endgroup$
                    – Wss
                    Oct 15 '18 at 4:44




                    $begingroup$
                    Nice approach but somewhat cheating to take $x=frac{c}{c+1}t$ at the beginning.
                    $endgroup$
                    – Wss
                    Oct 15 '18 at 4:44












                    $begingroup$
                    @Wss Because we know that the equality occurs for $x=frac{c}{c +1}$ and it would be much more better if it will happen for some variable is equal to $1$.
                    $endgroup$
                    – Michael Rozenberg
                    Oct 15 '18 at 4:47




                    $begingroup$
                    @Wss Because we know that the equality occurs for $x=frac{c}{c +1}$ and it would be much more better if it will happen for some variable is equal to $1$.
                    $endgroup$
                    – Michael Rozenberg
                    Oct 15 '18 at 4:47











                    1












                    $begingroup$

                    Assuming $$sqrt{x-x^2}+sqrt{cx-x^2}le sqrt{c}$$
                    Squaring we get



                    $$2sqrt{x-x^2}sqrt{cx-x^2}le 2x^2-x-cx+c$$
                    squaring again and factorizing we get $$(cx-c+x)^2geq 0$$ which is true.
                    Remark: We can only square the inequality if $$2x^2-x(c+1)+cgeq 0$$ this is true if $$0<cle 1$$ for $$0le xle c$$ or $$c>1$$ and $$0le xle 1$$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      @Sonnhard I think the right side of your second inequality can be negative. At least, you need to prove that it's not negative.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 9:27












                    • $begingroup$
                      This is true, the condition is complicated.
                      $endgroup$
                      – Dr. Sonnhard Graubner
                      Oct 14 '18 at 9:38










                    • $begingroup$
                      @Sonnhard But $c^2-6c+1$ can be positive, which says that the right side can be negative. No?
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 9:41










                    • $begingroup$
                      Yes with the inequalities $$x-x^2geq 0$$ and$$ cx-x^2geq 0$$ it simplifies to the given above
                      $endgroup$
                      – Dr. Sonnhard Graubner
                      Oct 14 '18 at 9:53












                    • $begingroup$
                      @Sonnhard But how did you got it? By the way, there is a very nice proof that $2x^2-(c+1)x+1>0$. Try to find it.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 9:59
















                    1












                    $begingroup$

                    Assuming $$sqrt{x-x^2}+sqrt{cx-x^2}le sqrt{c}$$
                    Squaring we get



                    $$2sqrt{x-x^2}sqrt{cx-x^2}le 2x^2-x-cx+c$$
                    squaring again and factorizing we get $$(cx-c+x)^2geq 0$$ which is true.
                    Remark: We can only square the inequality if $$2x^2-x(c+1)+cgeq 0$$ this is true if $$0<cle 1$$ for $$0le xle c$$ or $$c>1$$ and $$0le xle 1$$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      @Sonnhard I think the right side of your second inequality can be negative. At least, you need to prove that it's not negative.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 9:27












                    • $begingroup$
                      This is true, the condition is complicated.
                      $endgroup$
                      – Dr. Sonnhard Graubner
                      Oct 14 '18 at 9:38










                    • $begingroup$
                      @Sonnhard But $c^2-6c+1$ can be positive, which says that the right side can be negative. No?
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 9:41










                    • $begingroup$
                      Yes with the inequalities $$x-x^2geq 0$$ and$$ cx-x^2geq 0$$ it simplifies to the given above
                      $endgroup$
                      – Dr. Sonnhard Graubner
                      Oct 14 '18 at 9:53












                    • $begingroup$
                      @Sonnhard But how did you got it? By the way, there is a very nice proof that $2x^2-(c+1)x+1>0$. Try to find it.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 9:59














                    1












                    1








                    1





                    $begingroup$

                    Assuming $$sqrt{x-x^2}+sqrt{cx-x^2}le sqrt{c}$$
                    Squaring we get



                    $$2sqrt{x-x^2}sqrt{cx-x^2}le 2x^2-x-cx+c$$
                    squaring again and factorizing we get $$(cx-c+x)^2geq 0$$ which is true.
                    Remark: We can only square the inequality if $$2x^2-x(c+1)+cgeq 0$$ this is true if $$0<cle 1$$ for $$0le xle c$$ or $$c>1$$ and $$0le xle 1$$






                    share|cite|improve this answer











                    $endgroup$



                    Assuming $$sqrt{x-x^2}+sqrt{cx-x^2}le sqrt{c}$$
                    Squaring we get



                    $$2sqrt{x-x^2}sqrt{cx-x^2}le 2x^2-x-cx+c$$
                    squaring again and factorizing we get $$(cx-c+x)^2geq 0$$ which is true.
                    Remark: We can only square the inequality if $$2x^2-x(c+1)+cgeq 0$$ this is true if $$0<cle 1$$ for $$0le xle c$$ or $$c>1$$ and $$0le xle 1$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Oct 14 '18 at 9:51

























                    answered Oct 14 '18 at 9:25









                    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                    74.4k42865




                    74.4k42865












                    • $begingroup$
                      @Sonnhard I think the right side of your second inequality can be negative. At least, you need to prove that it's not negative.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 9:27












                    • $begingroup$
                      This is true, the condition is complicated.
                      $endgroup$
                      – Dr. Sonnhard Graubner
                      Oct 14 '18 at 9:38










                    • $begingroup$
                      @Sonnhard But $c^2-6c+1$ can be positive, which says that the right side can be negative. No?
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 9:41










                    • $begingroup$
                      Yes with the inequalities $$x-x^2geq 0$$ and$$ cx-x^2geq 0$$ it simplifies to the given above
                      $endgroup$
                      – Dr. Sonnhard Graubner
                      Oct 14 '18 at 9:53












                    • $begingroup$
                      @Sonnhard But how did you got it? By the way, there is a very nice proof that $2x^2-(c+1)x+1>0$. Try to find it.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 9:59


















                    • $begingroup$
                      @Sonnhard I think the right side of your second inequality can be negative. At least, you need to prove that it's not negative.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 9:27












                    • $begingroup$
                      This is true, the condition is complicated.
                      $endgroup$
                      – Dr. Sonnhard Graubner
                      Oct 14 '18 at 9:38










                    • $begingroup$
                      @Sonnhard But $c^2-6c+1$ can be positive, which says that the right side can be negative. No?
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 9:41










                    • $begingroup$
                      Yes with the inequalities $$x-x^2geq 0$$ and$$ cx-x^2geq 0$$ it simplifies to the given above
                      $endgroup$
                      – Dr. Sonnhard Graubner
                      Oct 14 '18 at 9:53












                    • $begingroup$
                      @Sonnhard But how did you got it? By the way, there is a very nice proof that $2x^2-(c+1)x+1>0$. Try to find it.
                      $endgroup$
                      – Michael Rozenberg
                      Oct 14 '18 at 9:59
















                    $begingroup$
                    @Sonnhard I think the right side of your second inequality can be negative. At least, you need to prove that it's not negative.
                    $endgroup$
                    – Michael Rozenberg
                    Oct 14 '18 at 9:27






                    $begingroup$
                    @Sonnhard I think the right side of your second inequality can be negative. At least, you need to prove that it's not negative.
                    $endgroup$
                    – Michael Rozenberg
                    Oct 14 '18 at 9:27














                    $begingroup$
                    This is true, the condition is complicated.
                    $endgroup$
                    – Dr. Sonnhard Graubner
                    Oct 14 '18 at 9:38




                    $begingroup$
                    This is true, the condition is complicated.
                    $endgroup$
                    – Dr. Sonnhard Graubner
                    Oct 14 '18 at 9:38












                    $begingroup$
                    @Sonnhard But $c^2-6c+1$ can be positive, which says that the right side can be negative. No?
                    $endgroup$
                    – Michael Rozenberg
                    Oct 14 '18 at 9:41




                    $begingroup$
                    @Sonnhard But $c^2-6c+1$ can be positive, which says that the right side can be negative. No?
                    $endgroup$
                    – Michael Rozenberg
                    Oct 14 '18 at 9:41












                    $begingroup$
                    Yes with the inequalities $$x-x^2geq 0$$ and$$ cx-x^2geq 0$$ it simplifies to the given above
                    $endgroup$
                    – Dr. Sonnhard Graubner
                    Oct 14 '18 at 9:53






                    $begingroup$
                    Yes with the inequalities $$x-x^2geq 0$$ and$$ cx-x^2geq 0$$ it simplifies to the given above
                    $endgroup$
                    – Dr. Sonnhard Graubner
                    Oct 14 '18 at 9:53














                    $begingroup$
                    @Sonnhard But how did you got it? By the way, there is a very nice proof that $2x^2-(c+1)x+1>0$. Try to find it.
                    $endgroup$
                    – Michael Rozenberg
                    Oct 14 '18 at 9:59




                    $begingroup$
                    @Sonnhard But how did you got it? By the way, there is a very nice proof that $2x^2-(c+1)x+1>0$. Try to find it.
                    $endgroup$
                    – Michael Rozenberg
                    Oct 14 '18 at 9:59


















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