Range of $sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha$ is












0












$begingroup$



If line makes an angle of $alpha,beta,gamma$ with positive $x,y$ and $z$



axis respectively. Then Range of



$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha$ is




Try: If line makes an angle of $alpha,beta,gamma$ with positive $x,y$ and



$z$ axis respectively. Then $cos^2 alpha+cos^2 beta+cos^2 gamma = 1$



means $sin^2 alpha+sin^2 beta+sin^2 gamma = 2.$



Using Cauchy Schwarz Inequality



$$(sin^2 alpha+sin^2 beta+sin^2 gamma )cdot (sin^2 beta+sin^2 gamma+sin^2 alpha)geq (sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha)^2$$



So $$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alphaleq 2$$



Now i did not understand how i find minimum value of $$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha$$



could some help me, Thanks










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Hint: What if the line is the $x$-axis? What are the angles in that case? What is the largest that any angle could possibly be? For the upper limit, can the value of $2$ be obtained?
    $endgroup$
    – Michael Burr
    Dec 4 '18 at 9:24












  • $begingroup$
    If line along $x$ axis . Then $alpha = 0^circ, beta = =gamma = 90^circ$. Michael i did not understand Why $2$ can not possible, please enlight me.
    $endgroup$
    – DXT
    Dec 4 '18 at 10:29






  • 1




    $begingroup$
    Yes it is. That's from the equality condition of Cauchy-Schwarz.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 4 '18 at 10:47










  • $begingroup$
    $2$ is attained, but you must prove it is attained (via an example).
    $endgroup$
    – Michael Burr
    Dec 4 '18 at 12:28
















0












$begingroup$



If line makes an angle of $alpha,beta,gamma$ with positive $x,y$ and $z$



axis respectively. Then Range of



$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha$ is




Try: If line makes an angle of $alpha,beta,gamma$ with positive $x,y$ and



$z$ axis respectively. Then $cos^2 alpha+cos^2 beta+cos^2 gamma = 1$



means $sin^2 alpha+sin^2 beta+sin^2 gamma = 2.$



Using Cauchy Schwarz Inequality



$$(sin^2 alpha+sin^2 beta+sin^2 gamma )cdot (sin^2 beta+sin^2 gamma+sin^2 alpha)geq (sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha)^2$$



So $$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alphaleq 2$$



Now i did not understand how i find minimum value of $$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha$$



could some help me, Thanks










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Hint: What if the line is the $x$-axis? What are the angles in that case? What is the largest that any angle could possibly be? For the upper limit, can the value of $2$ be obtained?
    $endgroup$
    – Michael Burr
    Dec 4 '18 at 9:24












  • $begingroup$
    If line along $x$ axis . Then $alpha = 0^circ, beta = =gamma = 90^circ$. Michael i did not understand Why $2$ can not possible, please enlight me.
    $endgroup$
    – DXT
    Dec 4 '18 at 10:29






  • 1




    $begingroup$
    Yes it is. That's from the equality condition of Cauchy-Schwarz.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 4 '18 at 10:47










  • $begingroup$
    $2$ is attained, but you must prove it is attained (via an example).
    $endgroup$
    – Michael Burr
    Dec 4 '18 at 12:28














0












0








0


1



$begingroup$



If line makes an angle of $alpha,beta,gamma$ with positive $x,y$ and $z$



axis respectively. Then Range of



$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha$ is




Try: If line makes an angle of $alpha,beta,gamma$ with positive $x,y$ and



$z$ axis respectively. Then $cos^2 alpha+cos^2 beta+cos^2 gamma = 1$



means $sin^2 alpha+sin^2 beta+sin^2 gamma = 2.$



Using Cauchy Schwarz Inequality



$$(sin^2 alpha+sin^2 beta+sin^2 gamma )cdot (sin^2 beta+sin^2 gamma+sin^2 alpha)geq (sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha)^2$$



So $$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alphaleq 2$$



Now i did not understand how i find minimum value of $$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha$$



could some help me, Thanks










share|cite|improve this question









$endgroup$





If line makes an angle of $alpha,beta,gamma$ with positive $x,y$ and $z$



axis respectively. Then Range of



$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha$ is




Try: If line makes an angle of $alpha,beta,gamma$ with positive $x,y$ and



$z$ axis respectively. Then $cos^2 alpha+cos^2 beta+cos^2 gamma = 1$



means $sin^2 alpha+sin^2 beta+sin^2 gamma = 2.$



Using Cauchy Schwarz Inequality



$$(sin^2 alpha+sin^2 beta+sin^2 gamma )cdot (sin^2 beta+sin^2 gamma+sin^2 alpha)geq (sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha)^2$$



So $$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alphaleq 2$$



Now i did not understand how i find minimum value of $$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha$$



could some help me, Thanks







inequality






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share|cite|improve this question











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asked Dec 4 '18 at 9:16









DXTDXT

5,5622630




5,5622630








  • 2




    $begingroup$
    Hint: What if the line is the $x$-axis? What are the angles in that case? What is the largest that any angle could possibly be? For the upper limit, can the value of $2$ be obtained?
    $endgroup$
    – Michael Burr
    Dec 4 '18 at 9:24












  • $begingroup$
    If line along $x$ axis . Then $alpha = 0^circ, beta = =gamma = 90^circ$. Michael i did not understand Why $2$ can not possible, please enlight me.
    $endgroup$
    – DXT
    Dec 4 '18 at 10:29






  • 1




    $begingroup$
    Yes it is. That's from the equality condition of Cauchy-Schwarz.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 4 '18 at 10:47










  • $begingroup$
    $2$ is attained, but you must prove it is attained (via an example).
    $endgroup$
    – Michael Burr
    Dec 4 '18 at 12:28














  • 2




    $begingroup$
    Hint: What if the line is the $x$-axis? What are the angles in that case? What is the largest that any angle could possibly be? For the upper limit, can the value of $2$ be obtained?
    $endgroup$
    – Michael Burr
    Dec 4 '18 at 9:24












  • $begingroup$
    If line along $x$ axis . Then $alpha = 0^circ, beta = =gamma = 90^circ$. Michael i did not understand Why $2$ can not possible, please enlight me.
    $endgroup$
    – DXT
    Dec 4 '18 at 10:29






  • 1




    $begingroup$
    Yes it is. That's from the equality condition of Cauchy-Schwarz.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 4 '18 at 10:47










  • $begingroup$
    $2$ is attained, but you must prove it is attained (via an example).
    $endgroup$
    – Michael Burr
    Dec 4 '18 at 12:28








2




2




$begingroup$
Hint: What if the line is the $x$-axis? What are the angles in that case? What is the largest that any angle could possibly be? For the upper limit, can the value of $2$ be obtained?
$endgroup$
– Michael Burr
Dec 4 '18 at 9:24






$begingroup$
Hint: What if the line is the $x$-axis? What are the angles in that case? What is the largest that any angle could possibly be? For the upper limit, can the value of $2$ be obtained?
$endgroup$
– Michael Burr
Dec 4 '18 at 9:24














$begingroup$
If line along $x$ axis . Then $alpha = 0^circ, beta = =gamma = 90^circ$. Michael i did not understand Why $2$ can not possible, please enlight me.
$endgroup$
– DXT
Dec 4 '18 at 10:29




$begingroup$
If line along $x$ axis . Then $alpha = 0^circ, beta = =gamma = 90^circ$. Michael i did not understand Why $2$ can not possible, please enlight me.
$endgroup$
– DXT
Dec 4 '18 at 10:29




1




1




$begingroup$
Yes it is. That's from the equality condition of Cauchy-Schwarz.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 10:47




$begingroup$
Yes it is. That's from the equality condition of Cauchy-Schwarz.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 10:47












$begingroup$
$2$ is attained, but you must prove it is attained (via an example).
$endgroup$
– Michael Burr
Dec 4 '18 at 12:28




$begingroup$
$2$ is attained, but you must prove it is attained (via an example).
$endgroup$
– Michael Burr
Dec 4 '18 at 12:28










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let $O(0,0,0)$, $A(sqrt{a},0,0),$ $B(0,sqrt{b},0)$ and $C(0,0,sqrt{c})$, where $a$, $b$ and $c$ are non-negatives such that $a^2+b^2+c^2neq0.$



Thus, we can assume that
$$sinalpha=sqrt{frac{b+c}{a+b+c}},$$
$$sinbeta=sqrt{frac{a+c}{a+b+c}}$$ and
$$singamma=sqrt{frac{a+b}{a+b+c}}.$$
Now, for $b=c=0$ we obtain:
$$sum_{cyc}sinalphasingamma=1.$$
We'll prove that it's a minimal value.



Indeed, we need to prove that
$$sum_{cyc}sqrt{frac{(a+b)(a+c)}{(a+b+c)^2}}geq1$$ or
$$sum_{cyc}sqrt{(a+b)(a+c)}geq a+b+c,$$ which is obvious because
$$sqrt{(a+b)(a+c)}geqsqrt{acdot a}=a.$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    For the upper limit of the range, you've proved that $2$ is an upper bound on the range, but not that $2$ actually equals the upper bound. To show that $2$ is the upper limit of the range, you must exhibit an example (or otherwise prove) that $2$ can be attained. In this case, the upper bound can be attained with the line in the direction $langle 1,1,1rangle$. In this case, the angle with all axes is the same, so, by the equality condition for Cauchy-Schwarz, you get the value of $2$ (or, alternately, use the cross product to calculate that the sine of the angle is $frac{sqrt{2}}{sqrt{3}}$ and calculate the desired value directly).



    For the lower bound, this occurs when the line is in the direction of one of the axes, e.g., $langle 1,0,0rangle$. In this case, one of the sines is zero and the other two are $1$. This leads to a value of $1$. This can be proved with a bit of multivariate calculus, if you wish, or a little figuring.



    Here's the sketch of how to argue that $1$ is the minimum value (without calculus): Observe that
    begin{align}
    (sinalpha+sinbeta+singamma)^2&=(sin^2alpha+sin^2beta+sin^2gamma)+2(sinalphasinbeta+sinalphasingamma+sinbetasingamma)\
    &=2+2(sinalphasinbeta+sinalphasingamma+sinbetasingamma).
    end{align}

    Therefore, since everything is nonnegative, minimizing the desired quantity is the same as minimizing $sinalpha+sinbeta+singamma$. Since each sine is between $0$ and $1$, we know that $sin^2alphaleqsinalpha$, $sin^2betaleqsinbeta$, and $sin^2gammaleqsingamma$. Therefore,
    $$
    2=sin^2alpha+sin^2beta+sin^2gammaleq sinalpha+sinbeta+singamma.
    $$

    The minimum is attained for the example above.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Let $O(0,0,0)$, $A(sqrt{a},0,0),$ $B(0,sqrt{b},0)$ and $C(0,0,sqrt{c})$, where $a$, $b$ and $c$ are non-negatives such that $a^2+b^2+c^2neq0.$



      Thus, we can assume that
      $$sinalpha=sqrt{frac{b+c}{a+b+c}},$$
      $$sinbeta=sqrt{frac{a+c}{a+b+c}}$$ and
      $$singamma=sqrt{frac{a+b}{a+b+c}}.$$
      Now, for $b=c=0$ we obtain:
      $$sum_{cyc}sinalphasingamma=1.$$
      We'll prove that it's a minimal value.



      Indeed, we need to prove that
      $$sum_{cyc}sqrt{frac{(a+b)(a+c)}{(a+b+c)^2}}geq1$$ or
      $$sum_{cyc}sqrt{(a+b)(a+c)}geq a+b+c,$$ which is obvious because
      $$sqrt{(a+b)(a+c)}geqsqrt{acdot a}=a.$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Let $O(0,0,0)$, $A(sqrt{a},0,0),$ $B(0,sqrt{b},0)$ and $C(0,0,sqrt{c})$, where $a$, $b$ and $c$ are non-negatives such that $a^2+b^2+c^2neq0.$



        Thus, we can assume that
        $$sinalpha=sqrt{frac{b+c}{a+b+c}},$$
        $$sinbeta=sqrt{frac{a+c}{a+b+c}}$$ and
        $$singamma=sqrt{frac{a+b}{a+b+c}}.$$
        Now, for $b=c=0$ we obtain:
        $$sum_{cyc}sinalphasingamma=1.$$
        We'll prove that it's a minimal value.



        Indeed, we need to prove that
        $$sum_{cyc}sqrt{frac{(a+b)(a+c)}{(a+b+c)^2}}geq1$$ or
        $$sum_{cyc}sqrt{(a+b)(a+c)}geq a+b+c,$$ which is obvious because
        $$sqrt{(a+b)(a+c)}geqsqrt{acdot a}=a.$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Let $O(0,0,0)$, $A(sqrt{a},0,0),$ $B(0,sqrt{b},0)$ and $C(0,0,sqrt{c})$, where $a$, $b$ and $c$ are non-negatives such that $a^2+b^2+c^2neq0.$



          Thus, we can assume that
          $$sinalpha=sqrt{frac{b+c}{a+b+c}},$$
          $$sinbeta=sqrt{frac{a+c}{a+b+c}}$$ and
          $$singamma=sqrt{frac{a+b}{a+b+c}}.$$
          Now, for $b=c=0$ we obtain:
          $$sum_{cyc}sinalphasingamma=1.$$
          We'll prove that it's a minimal value.



          Indeed, we need to prove that
          $$sum_{cyc}sqrt{frac{(a+b)(a+c)}{(a+b+c)^2}}geq1$$ or
          $$sum_{cyc}sqrt{(a+b)(a+c)}geq a+b+c,$$ which is obvious because
          $$sqrt{(a+b)(a+c)}geqsqrt{acdot a}=a.$$






          share|cite|improve this answer











          $endgroup$



          Let $O(0,0,0)$, $A(sqrt{a},0,0),$ $B(0,sqrt{b},0)$ and $C(0,0,sqrt{c})$, where $a$, $b$ and $c$ are non-negatives such that $a^2+b^2+c^2neq0.$



          Thus, we can assume that
          $$sinalpha=sqrt{frac{b+c}{a+b+c}},$$
          $$sinbeta=sqrt{frac{a+c}{a+b+c}}$$ and
          $$singamma=sqrt{frac{a+b}{a+b+c}}.$$
          Now, for $b=c=0$ we obtain:
          $$sum_{cyc}sinalphasingamma=1.$$
          We'll prove that it's a minimal value.



          Indeed, we need to prove that
          $$sum_{cyc}sqrt{frac{(a+b)(a+c)}{(a+b+c)^2}}geq1$$ or
          $$sum_{cyc}sqrt{(a+b)(a+c)}geq a+b+c,$$ which is obvious because
          $$sqrt{(a+b)(a+c)}geqsqrt{acdot a}=a.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 5 '18 at 5:31

























          answered Dec 4 '18 at 22:21









          Michael RozenbergMichael Rozenberg

          98.7k1590189




          98.7k1590189























              1












              $begingroup$

              For the upper limit of the range, you've proved that $2$ is an upper bound on the range, but not that $2$ actually equals the upper bound. To show that $2$ is the upper limit of the range, you must exhibit an example (or otherwise prove) that $2$ can be attained. In this case, the upper bound can be attained with the line in the direction $langle 1,1,1rangle$. In this case, the angle with all axes is the same, so, by the equality condition for Cauchy-Schwarz, you get the value of $2$ (or, alternately, use the cross product to calculate that the sine of the angle is $frac{sqrt{2}}{sqrt{3}}$ and calculate the desired value directly).



              For the lower bound, this occurs when the line is in the direction of one of the axes, e.g., $langle 1,0,0rangle$. In this case, one of the sines is zero and the other two are $1$. This leads to a value of $1$. This can be proved with a bit of multivariate calculus, if you wish, or a little figuring.



              Here's the sketch of how to argue that $1$ is the minimum value (without calculus): Observe that
              begin{align}
              (sinalpha+sinbeta+singamma)^2&=(sin^2alpha+sin^2beta+sin^2gamma)+2(sinalphasinbeta+sinalphasingamma+sinbetasingamma)\
              &=2+2(sinalphasinbeta+sinalphasingamma+sinbetasingamma).
              end{align}

              Therefore, since everything is nonnegative, minimizing the desired quantity is the same as minimizing $sinalpha+sinbeta+singamma$. Since each sine is between $0$ and $1$, we know that $sin^2alphaleqsinalpha$, $sin^2betaleqsinbeta$, and $sin^2gammaleqsingamma$. Therefore,
              $$
              2=sin^2alpha+sin^2beta+sin^2gammaleq sinalpha+sinbeta+singamma.
              $$

              The minimum is attained for the example above.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                For the upper limit of the range, you've proved that $2$ is an upper bound on the range, but not that $2$ actually equals the upper bound. To show that $2$ is the upper limit of the range, you must exhibit an example (or otherwise prove) that $2$ can be attained. In this case, the upper bound can be attained with the line in the direction $langle 1,1,1rangle$. In this case, the angle with all axes is the same, so, by the equality condition for Cauchy-Schwarz, you get the value of $2$ (or, alternately, use the cross product to calculate that the sine of the angle is $frac{sqrt{2}}{sqrt{3}}$ and calculate the desired value directly).



                For the lower bound, this occurs when the line is in the direction of one of the axes, e.g., $langle 1,0,0rangle$. In this case, one of the sines is zero and the other two are $1$. This leads to a value of $1$. This can be proved with a bit of multivariate calculus, if you wish, or a little figuring.



                Here's the sketch of how to argue that $1$ is the minimum value (without calculus): Observe that
                begin{align}
                (sinalpha+sinbeta+singamma)^2&=(sin^2alpha+sin^2beta+sin^2gamma)+2(sinalphasinbeta+sinalphasingamma+sinbetasingamma)\
                &=2+2(sinalphasinbeta+sinalphasingamma+sinbetasingamma).
                end{align}

                Therefore, since everything is nonnegative, minimizing the desired quantity is the same as minimizing $sinalpha+sinbeta+singamma$. Since each sine is between $0$ and $1$, we know that $sin^2alphaleqsinalpha$, $sin^2betaleqsinbeta$, and $sin^2gammaleqsingamma$. Therefore,
                $$
                2=sin^2alpha+sin^2beta+sin^2gammaleq sinalpha+sinbeta+singamma.
                $$

                The minimum is attained for the example above.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  For the upper limit of the range, you've proved that $2$ is an upper bound on the range, but not that $2$ actually equals the upper bound. To show that $2$ is the upper limit of the range, you must exhibit an example (or otherwise prove) that $2$ can be attained. In this case, the upper bound can be attained with the line in the direction $langle 1,1,1rangle$. In this case, the angle with all axes is the same, so, by the equality condition for Cauchy-Schwarz, you get the value of $2$ (or, alternately, use the cross product to calculate that the sine of the angle is $frac{sqrt{2}}{sqrt{3}}$ and calculate the desired value directly).



                  For the lower bound, this occurs when the line is in the direction of one of the axes, e.g., $langle 1,0,0rangle$. In this case, one of the sines is zero and the other two are $1$. This leads to a value of $1$. This can be proved with a bit of multivariate calculus, if you wish, or a little figuring.



                  Here's the sketch of how to argue that $1$ is the minimum value (without calculus): Observe that
                  begin{align}
                  (sinalpha+sinbeta+singamma)^2&=(sin^2alpha+sin^2beta+sin^2gamma)+2(sinalphasinbeta+sinalphasingamma+sinbetasingamma)\
                  &=2+2(sinalphasinbeta+sinalphasingamma+sinbetasingamma).
                  end{align}

                  Therefore, since everything is nonnegative, minimizing the desired quantity is the same as minimizing $sinalpha+sinbeta+singamma$. Since each sine is between $0$ and $1$, we know that $sin^2alphaleqsinalpha$, $sin^2betaleqsinbeta$, and $sin^2gammaleqsingamma$. Therefore,
                  $$
                  2=sin^2alpha+sin^2beta+sin^2gammaleq sinalpha+sinbeta+singamma.
                  $$

                  The minimum is attained for the example above.






                  share|cite|improve this answer









                  $endgroup$



                  For the upper limit of the range, you've proved that $2$ is an upper bound on the range, but not that $2$ actually equals the upper bound. To show that $2$ is the upper limit of the range, you must exhibit an example (or otherwise prove) that $2$ can be attained. In this case, the upper bound can be attained with the line in the direction $langle 1,1,1rangle$. In this case, the angle with all axes is the same, so, by the equality condition for Cauchy-Schwarz, you get the value of $2$ (or, alternately, use the cross product to calculate that the sine of the angle is $frac{sqrt{2}}{sqrt{3}}$ and calculate the desired value directly).



                  For the lower bound, this occurs when the line is in the direction of one of the axes, e.g., $langle 1,0,0rangle$. In this case, one of the sines is zero and the other two are $1$. This leads to a value of $1$. This can be proved with a bit of multivariate calculus, if you wish, or a little figuring.



                  Here's the sketch of how to argue that $1$ is the minimum value (without calculus): Observe that
                  begin{align}
                  (sinalpha+sinbeta+singamma)^2&=(sin^2alpha+sin^2beta+sin^2gamma)+2(sinalphasinbeta+sinalphasingamma+sinbetasingamma)\
                  &=2+2(sinalphasinbeta+sinalphasingamma+sinbetasingamma).
                  end{align}

                  Therefore, since everything is nonnegative, minimizing the desired quantity is the same as minimizing $sinalpha+sinbeta+singamma$. Since each sine is between $0$ and $1$, we know that $sin^2alphaleqsinalpha$, $sin^2betaleqsinbeta$, and $sin^2gammaleqsingamma$. Therefore,
                  $$
                  2=sin^2alpha+sin^2beta+sin^2gammaleq sinalpha+sinbeta+singamma.
                  $$

                  The minimum is attained for the example above.







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                  answered Dec 4 '18 at 12:50









                  Michael BurrMichael Burr

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