Simple two variable am-gm inequality











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Given $x,y in Bbb{R}$, show that:$$x^2+y^2+1ge xy+y+x $$
I tried using the fact that $x^2+y^2 ge 2xy$ But then I'm not sure how to go on, Also tried factoring but didn't help much, also tried substituting $frac{x^2+y^2}{2}$ instead of $xy$ but that gave me the same result of the first substitution, i.e. $xy+1ge x+y$



This inequality seems very easy, I'm feeling dumb for not having solved it yet






inequality a.m.-g.m.-inequality







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  • Fix $y$, differentiate w.r.t. $x$, etc
    – mathworker21
    1 hour ago










  • Possible duplicate of How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$? – Also: math.stackexchange.com/q/772220/42969
    – Martin R
    1 hour ago












  • @MartinR I just didn't see them being equivalent, I well new that one inequality
    – Spasoje Durovic
    1 hour ago










  • @Spasoje Durovic Substitute $c=1$.
    – Michael Rozenberg
    57 mins ago















up vote
2
down vote

favorite












Given $x,y in Bbb{R}$, show that:$$x^2+y^2+1ge xy+y+x $$
I tried using the fact that $x^2+y^2 ge 2xy$ But then I'm not sure how to go on, Also tried factoring but didn't help much, also tried substituting $frac{x^2+y^2}{2}$ instead of $xy$ but that gave me the same result of the first substitution, i.e. $xy+1ge x+y$



This inequality seems very easy, I'm feeling dumb for not having solved it yet






inequality a.m.-g.m.-inequality







share|cite|improve this question






















  • Fix $y$, differentiate w.r.t. $x$, etc
    – mathworker21
    1 hour ago










  • Possible duplicate of How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$? – Also: math.stackexchange.com/q/772220/42969
    – Martin R
    1 hour ago












  • @MartinR I just didn't see them being equivalent, I well new that one inequality
    – Spasoje Durovic
    1 hour ago










  • @Spasoje Durovic Substitute $c=1$.
    – Michael Rozenberg
    57 mins ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Given $x,y in Bbb{R}$, show that:$$x^2+y^2+1ge xy+y+x $$
I tried using the fact that $x^2+y^2 ge 2xy$ But then I'm not sure how to go on, Also tried factoring but didn't help much, also tried substituting $frac{x^2+y^2}{2}$ instead of $xy$ but that gave me the same result of the first substitution, i.e. $xy+1ge x+y$



This inequality seems very easy, I'm feeling dumb for not having solved it yet






inequality a.m.-g.m.-inequality







share|cite|improve this question













Given $x,y in Bbb{R}$, show that:$$x^2+y^2+1ge xy+y+x $$
I tried using the fact that $x^2+y^2 ge 2xy$ But then I'm not sure how to go on, Also tried factoring but didn't help much, also tried substituting $frac{x^2+y^2}{2}$ instead of $xy$ but that gave me the same result of the first substitution, i.e. $xy+1ge x+y$



This inequality seems very easy, I'm feeling dumb for not having solved it yet






inequality a.m.-g.m.-inequality




inequality 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










asked 1 hour ago









Spasoje Durovic

1809




1809












  • Fix $y$, differentiate w.r.t. $x$, etc
    – mathworker21
    1 hour ago










  • Possible duplicate of How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$? – Also: math.stackexchange.com/q/772220/42969
    – Martin R
    1 hour ago












  • @MartinR I just didn't see them being equivalent, I well new that one inequality
    – Spasoje Durovic
    1 hour ago










  • @Spasoje Durovic Substitute $c=1$.
    – Michael Rozenberg
    57 mins ago


















  • Fix $y$, differentiate w.r.t. $x$, etc
    – mathworker21
    1 hour ago










  • Possible duplicate of How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$? – Also: math.stackexchange.com/q/772220/42969
    – Martin R
    1 hour ago












  • @MartinR I just didn't see them being equivalent, I well new that one inequality
    – Spasoje Durovic
    1 hour ago










  • @Spasoje Durovic Substitute $c=1$.
    – Michael Rozenberg
    57 mins ago
















Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
1 hour ago




Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
1 hour ago












Possible duplicate of How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$? – Also: math.stackexchange.com/q/772220/42969
– Martin R
1 hour ago






Possible duplicate of How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$? – Also: math.stackexchange.com/q/772220/42969
– Martin R
1 hour ago














@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
1 hour ago




@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
1 hour ago












@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
57 mins ago




@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
57 mins ago










3 Answers
3






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up vote
7
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accepted










Since $$a^2-2ab+b^2 = (a-b)^2geq 0implies a^2+b^2geq 2ab$$ we have $$ x^2+y^2geq 2xy$$
$$x^2+1geq 2x$$
$$y^2+1geq 2y$$
Now add all these...






share|cite|improve this answer




























    up vote
    5
    down vote













    Hint: Use that $$x^2+y^2+z^2geq xy+yz+zx$$ holds for all real numbers $$x,y,z$$
    this is $$(x-y)^2+(y-z)^2+(z-x)^2geq 0$$






    share|cite|improve this answer




























      up vote
      2
      down vote













      We need to prove that
      $$y^2-(x+1)y+x^2-x+1geq0,$$ for which it's enough to prove that
      $$(x+1)^2-4(x^2-x+1)leq0$$ or $$(x-1)^2geq0.$$






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        7
        down vote



        accepted










        Since $$a^2-2ab+b^2 = (a-b)^2geq 0implies a^2+b^2geq 2ab$$ we have $$ x^2+y^2geq 2xy$$
        $$x^2+1geq 2x$$
        $$y^2+1geq 2y$$
        Now add all these...






        share|cite|improve this answer

























          up vote
          7
          down vote



          accepted










          Since $$a^2-2ab+b^2 = (a-b)^2geq 0implies a^2+b^2geq 2ab$$ we have $$ x^2+y^2geq 2xy$$
          $$x^2+1geq 2x$$
          $$y^2+1geq 2y$$
          Now add all these...






          share|cite|improve this answer























            up vote
            7
            down vote



            accepted







            up vote
            7
            down vote



            accepted






            Since $$a^2-2ab+b^2 = (a-b)^2geq 0implies a^2+b^2geq 2ab$$ we have $$ x^2+y^2geq 2xy$$
            $$x^2+1geq 2x$$
            $$y^2+1geq 2y$$
            Now add all these...






            share|cite|improve this answer












            Since $$a^2-2ab+b^2 = (a-b)^2geq 0implies a^2+b^2geq 2ab$$ we have $$ x^2+y^2geq 2xy$$
            $$x^2+1geq 2x$$
            $$y^2+1geq 2y$$
            Now add all these...







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            greedoid

            36.6k114592




            36.6k114592






















                up vote
                5
                down vote













                Hint: Use that $$x^2+y^2+z^2geq xy+yz+zx$$ holds for all real numbers $$x,y,z$$
                this is $$(x-y)^2+(y-z)^2+(z-x)^2geq 0$$






                share|cite|improve this answer

























                  up vote
                  5
                  down vote













                  Hint: Use that $$x^2+y^2+z^2geq xy+yz+zx$$ holds for all real numbers $$x,y,z$$
                  this is $$(x-y)^2+(y-z)^2+(z-x)^2geq 0$$






                  share|cite|improve this answer























                    up vote
                    5
                    down vote










                    up vote
                    5
                    down vote









                    Hint: Use that $$x^2+y^2+z^2geq xy+yz+zx$$ holds for all real numbers $$x,y,z$$
                    this is $$(x-y)^2+(y-z)^2+(z-x)^2geq 0$$






                    share|cite|improve this answer












                    Hint: Use that $$x^2+y^2+z^2geq xy+yz+zx$$ holds for all real numbers $$x,y,z$$
                    this is $$(x-y)^2+(y-z)^2+(z-x)^2geq 0$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Dr. Sonnhard Graubner

                    72.4k32865




                    72.4k32865






















                        up vote
                        2
                        down vote













                        We need to prove that
                        $$y^2-(x+1)y+x^2-x+1geq0,$$ for which it's enough to prove that
                        $$(x+1)^2-4(x^2-x+1)leq0$$ or $$(x-1)^2geq0.$$






                        share|cite|improve this answer

























                          up vote
                          2
                          down vote













                          We need to prove that
                          $$y^2-(x+1)y+x^2-x+1geq0,$$ for which it's enough to prove that
                          $$(x+1)^2-4(x^2-x+1)leq0$$ or $$(x-1)^2geq0.$$






                          share|cite|improve this answer























                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            We need to prove that
                            $$y^2-(x+1)y+x^2-x+1geq0,$$ for which it's enough to prove that
                            $$(x+1)^2-4(x^2-x+1)leq0$$ or $$(x-1)^2geq0.$$






                            share|cite|improve this answer












                            We need to prove that
                            $$y^2-(x+1)y+x^2-x+1geq0,$$ for which it's enough to prove that
                            $$(x+1)^2-4(x^2-x+1)leq0$$ or $$(x-1)^2geq0.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            Michael Rozenberg

                            95k1588183




                            95k1588183






























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