Find maximum of $cos(x)cos(y)cos(z)$ when $x + y + z = frac{pi}{2}$ and $x, y, z > 0$












2












$begingroup$


Find maximum value of P = $cos(x)cos(y)cos(z)$, given that $x + y + z = frac{pi}{2}$ and $x, y, z > 0$.



Effort 1. I drew a quarter-circle, divided the square angle into three parts, and attempted to derive the expression, but it went to nowhere.






real-analysis trigonometry optimization maxima-minima jensen-inequality







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$endgroup$

















    2












    $begingroup$


    Find maximum value of P = $cos(x)cos(y)cos(z)$, given that $x + y + z = frac{pi}{2}$ and $x, y, z > 0$.



    Effort 1. I drew a quarter-circle, divided the square angle into three parts, and attempted to derive the expression, but it went to nowhere.






    real-analysis trigonometry optimization maxima-minima jensen-inequality







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Find maximum value of P = $cos(x)cos(y)cos(z)$, given that $x + y + z = frac{pi}{2}$ and $x, y, z > 0$.



      Effort 1. I drew a quarter-circle, divided the square angle into three parts, and attempted to derive the expression, but it went to nowhere.






      real-analysis trigonometry optimization maxima-minima jensen-inequality







      share|cite|improve this question











      $endgroup$




      Find maximum value of P = $cos(x)cos(y)cos(z)$, given that $x + y + z = frac{pi}{2}$ and $x, y, z > 0$.



      Effort 1. I drew a quarter-circle, divided the square angle into three parts, and attempted to derive the expression, but it went to nowhere.






      real-analysis trigonometry optimization maxima-minima jensen-inequality




      real-analysis trigonometry optimization maxima-minima jensen-inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 2 at 12:39









      Michael Rozenberg

      110k1896201




      110k1896201










      asked Feb 2 at 5:49









      KiWiJuiceKiWiJuice

      186




      186






















          2 Answers
          2






          active

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          5












          $begingroup$

          Let $f(x)=lncos{x}.$



          Thus, $$f''(x)=-left(tan{x}right)'=-frac{1}{cos^2x}<0.$$ Thus, $f$ is a concave function and since $e^x$ increases, by Jensen we obtain:
          $$cos{x}cos{y}cos{z}=e^{sumlimits_{cyc}lncos{x}}leq e^{3lncosfrac{x+y+z}{3}}=frac{3sqrt3}{8}.$$
          The equality occurs for $x=y=z=frac{pi}{6},$ which says that we got a maximal value.






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            A geometric approach:



            Consider an acute triangle $ABC$ with angles labeled by their vertices, and set $x=frac{pi}{2}-A$, $y=frac{pi}{2}-B$, $z=frac{pi}{2}-C$. Since $A+B+C=pi$, $x+y+z=frac{pi}{2}$. The acute triangle restriction ensures that $x,y,z>0$, and the restrictions are the same. In terms of $A,B,C$, the quantity we seek to maximize is $sin Asin Bsin C$.



            Now, we get clever. The area of triangle $ABC$ is equal to $2R^2sin Asin Bsin C$, where $R$ is the circumradius of the triangle. Why? Because $2Rsin A=a$ and $2Rsin B=b$, we can reduce to the more familiar area formula $frac12absin C$. Maximizing $sin Asin Bsin C$ is now equivalent to maximizing the area of an acute triangle inscribed in a fixed circle $Gamma$.



            And to do that, let's loosen things a bit, and instead maximize the area of any triangle inscribed in $Gamma$. If we fix two points $A$ and $B$ on the circle, what should $C$ be to maximize the area? It should be as far away from the line $AB$ as possible, namely one of the endpoints of the diameter of $Gamma$ that is the perpendicular bisector of $AB$. Comparing the two endpoints, the farther one is the one with the acute angle (or, if they're both right angles, both are equally far), so the area maximum with $A$ and $B$ fixed has $a=b$ and angle $C$ acute. Repeat for the other two pairs of vertices, and the maximum must come when $a=b=c$ and the triangle is equilateral. That's $60^circ$ angles $A,B,C$, for $x=y=z=frac{pi}{6}$ and $cos xcos ycos z = sin Asin Bsin C=left(frac{sqrt{3}}{2}right)^3=frac{3sqrt{3}}{8}$.



            Now, that's a little unsatisfying logically - we proved that the maximum, if it exists, must be at an equilateral triangle. To patch this, we prove that there is a maximum. The area is a continuous function from $Gamma^3$ (the positions of the three points) to $mathbb{R}$. That's a compact domain, so the extreme value theorem applies and there is a maximum. Done.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
              $endgroup$
              – KiWiJuice
              Mar 14 at 7:44










            • $begingroup$
              While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
              $endgroup$
              – jmerry
              Mar 14 at 7:48












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            2 Answers
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            5












            $begingroup$

            Let $f(x)=lncos{x}.$



            Thus, $$f''(x)=-left(tan{x}right)'=-frac{1}{cos^2x}<0.$$ Thus, $f$ is a concave function and since $e^x$ increases, by Jensen we obtain:
            $$cos{x}cos{y}cos{z}=e^{sumlimits_{cyc}lncos{x}}leq e^{3lncosfrac{x+y+z}{3}}=frac{3sqrt3}{8}.$$
            The equality occurs for $x=y=z=frac{pi}{6},$ which says that we got a maximal value.






            share|cite|improve this answer











            $endgroup$


















              5












              $begingroup$

              Let $f(x)=lncos{x}.$



              Thus, $$f''(x)=-left(tan{x}right)'=-frac{1}{cos^2x}<0.$$ Thus, $f$ is a concave function and since $e^x$ increases, by Jensen we obtain:
              $$cos{x}cos{y}cos{z}=e^{sumlimits_{cyc}lncos{x}}leq e^{3lncosfrac{x+y+z}{3}}=frac{3sqrt3}{8}.$$
              The equality occurs for $x=y=z=frac{pi}{6},$ which says that we got a maximal value.






              share|cite|improve this answer











              $endgroup$
















                5












                5








                5





                $begingroup$

                Let $f(x)=lncos{x}.$



                Thus, $$f''(x)=-left(tan{x}right)'=-frac{1}{cos^2x}<0.$$ Thus, $f$ is a concave function and since $e^x$ increases, by Jensen we obtain:
                $$cos{x}cos{y}cos{z}=e^{sumlimits_{cyc}lncos{x}}leq e^{3lncosfrac{x+y+z}{3}}=frac{3sqrt3}{8}.$$
                The equality occurs for $x=y=z=frac{pi}{6},$ which says that we got a maximal value.






                share|cite|improve this answer











                $endgroup$



                Let $f(x)=lncos{x}.$



                Thus, $$f''(x)=-left(tan{x}right)'=-frac{1}{cos^2x}<0.$$ Thus, $f$ is a concave function and since $e^x$ increases, by Jensen we obtain:
                $$cos{x}cos{y}cos{z}=e^{sumlimits_{cyc}lncos{x}}leq e^{3lncosfrac{x+y+z}{3}}=frac{3sqrt3}{8}.$$
                The equality occurs for $x=y=z=frac{pi}{6},$ which says that we got a maximal value.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 2 at 5:57

























                answered Feb 2 at 5:52









                Michael RozenbergMichael Rozenberg

                110k1896201




                110k1896201























                    2












                    $begingroup$

                    A geometric approach:



                    Consider an acute triangle $ABC$ with angles labeled by their vertices, and set $x=frac{pi}{2}-A$, $y=frac{pi}{2}-B$, $z=frac{pi}{2}-C$. Since $A+B+C=pi$, $x+y+z=frac{pi}{2}$. The acute triangle restriction ensures that $x,y,z>0$, and the restrictions are the same. In terms of $A,B,C$, the quantity we seek to maximize is $sin Asin Bsin C$.



                    Now, we get clever. The area of triangle $ABC$ is equal to $2R^2sin Asin Bsin C$, where $R$ is the circumradius of the triangle. Why? Because $2Rsin A=a$ and $2Rsin B=b$, we can reduce to the more familiar area formula $frac12absin C$. Maximizing $sin Asin Bsin C$ is now equivalent to maximizing the area of an acute triangle inscribed in a fixed circle $Gamma$.



                    And to do that, let's loosen things a bit, and instead maximize the area of any triangle inscribed in $Gamma$. If we fix two points $A$ and $B$ on the circle, what should $C$ be to maximize the area? It should be as far away from the line $AB$ as possible, namely one of the endpoints of the diameter of $Gamma$ that is the perpendicular bisector of $AB$. Comparing the two endpoints, the farther one is the one with the acute angle (or, if they're both right angles, both are equally far), so the area maximum with $A$ and $B$ fixed has $a=b$ and angle $C$ acute. Repeat for the other two pairs of vertices, and the maximum must come when $a=b=c$ and the triangle is equilateral. That's $60^circ$ angles $A,B,C$, for $x=y=z=frac{pi}{6}$ and $cos xcos ycos z = sin Asin Bsin C=left(frac{sqrt{3}}{2}right)^3=frac{3sqrt{3}}{8}$.



                    Now, that's a little unsatisfying logically - we proved that the maximum, if it exists, must be at an equilateral triangle. To patch this, we prove that there is a maximum. The area is a continuous function from $Gamma^3$ (the positions of the three points) to $mathbb{R}$. That's a compact domain, so the extreme value theorem applies and there is a maximum. Done.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
                      $endgroup$
                      – KiWiJuice
                      Mar 14 at 7:44










                    • $begingroup$
                      While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
                      $endgroup$
                      – jmerry
                      Mar 14 at 7:48
















                    2












                    $begingroup$

                    A geometric approach:



                    Consider an acute triangle $ABC$ with angles labeled by their vertices, and set $x=frac{pi}{2}-A$, $y=frac{pi}{2}-B$, $z=frac{pi}{2}-C$. Since $A+B+C=pi$, $x+y+z=frac{pi}{2}$. The acute triangle restriction ensures that $x,y,z>0$, and the restrictions are the same. In terms of $A,B,C$, the quantity we seek to maximize is $sin Asin Bsin C$.



                    Now, we get clever. The area of triangle $ABC$ is equal to $2R^2sin Asin Bsin C$, where $R$ is the circumradius of the triangle. Why? Because $2Rsin A=a$ and $2Rsin B=b$, we can reduce to the more familiar area formula $frac12absin C$. Maximizing $sin Asin Bsin C$ is now equivalent to maximizing the area of an acute triangle inscribed in a fixed circle $Gamma$.



                    And to do that, let's loosen things a bit, and instead maximize the area of any triangle inscribed in $Gamma$. If we fix two points $A$ and $B$ on the circle, what should $C$ be to maximize the area? It should be as far away from the line $AB$ as possible, namely one of the endpoints of the diameter of $Gamma$ that is the perpendicular bisector of $AB$. Comparing the two endpoints, the farther one is the one with the acute angle (or, if they're both right angles, both are equally far), so the area maximum with $A$ and $B$ fixed has $a=b$ and angle $C$ acute. Repeat for the other two pairs of vertices, and the maximum must come when $a=b=c$ and the triangle is equilateral. That's $60^circ$ angles $A,B,C$, for $x=y=z=frac{pi}{6}$ and $cos xcos ycos z = sin Asin Bsin C=left(frac{sqrt{3}}{2}right)^3=frac{3sqrt{3}}{8}$.



                    Now, that's a little unsatisfying logically - we proved that the maximum, if it exists, must be at an equilateral triangle. To patch this, we prove that there is a maximum. The area is a continuous function from $Gamma^3$ (the positions of the three points) to $mathbb{R}$. That's a compact domain, so the extreme value theorem applies and there is a maximum. Done.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
                      $endgroup$
                      – KiWiJuice
                      Mar 14 at 7:44










                    • $begingroup$
                      While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
                      $endgroup$
                      – jmerry
                      Mar 14 at 7:48














                    2












                    2








                    2





                    $begingroup$

                    A geometric approach:



                    Consider an acute triangle $ABC$ with angles labeled by their vertices, and set $x=frac{pi}{2}-A$, $y=frac{pi}{2}-B$, $z=frac{pi}{2}-C$. Since $A+B+C=pi$, $x+y+z=frac{pi}{2}$. The acute triangle restriction ensures that $x,y,z>0$, and the restrictions are the same. In terms of $A,B,C$, the quantity we seek to maximize is $sin Asin Bsin C$.



                    Now, we get clever. The area of triangle $ABC$ is equal to $2R^2sin Asin Bsin C$, where $R$ is the circumradius of the triangle. Why? Because $2Rsin A=a$ and $2Rsin B=b$, we can reduce to the more familiar area formula $frac12absin C$. Maximizing $sin Asin Bsin C$ is now equivalent to maximizing the area of an acute triangle inscribed in a fixed circle $Gamma$.



                    And to do that, let's loosen things a bit, and instead maximize the area of any triangle inscribed in $Gamma$. If we fix two points $A$ and $B$ on the circle, what should $C$ be to maximize the area? It should be as far away from the line $AB$ as possible, namely one of the endpoints of the diameter of $Gamma$ that is the perpendicular bisector of $AB$. Comparing the two endpoints, the farther one is the one with the acute angle (or, if they're both right angles, both are equally far), so the area maximum with $A$ and $B$ fixed has $a=b$ and angle $C$ acute. Repeat for the other two pairs of vertices, and the maximum must come when $a=b=c$ and the triangle is equilateral. That's $60^circ$ angles $A,B,C$, for $x=y=z=frac{pi}{6}$ and $cos xcos ycos z = sin Asin Bsin C=left(frac{sqrt{3}}{2}right)^3=frac{3sqrt{3}}{8}$.



                    Now, that's a little unsatisfying logically - we proved that the maximum, if it exists, must be at an equilateral triangle. To patch this, we prove that there is a maximum. The area is a continuous function from $Gamma^3$ (the positions of the three points) to $mathbb{R}$. That's a compact domain, so the extreme value theorem applies and there is a maximum. Done.






                    share|cite|improve this answer









                    $endgroup$



                    A geometric approach:



                    Consider an acute triangle $ABC$ with angles labeled by their vertices, and set $x=frac{pi}{2}-A$, $y=frac{pi}{2}-B$, $z=frac{pi}{2}-C$. Since $A+B+C=pi$, $x+y+z=frac{pi}{2}$. The acute triangle restriction ensures that $x,y,z>0$, and the restrictions are the same. In terms of $A,B,C$, the quantity we seek to maximize is $sin Asin Bsin C$.



                    Now, we get clever. The area of triangle $ABC$ is equal to $2R^2sin Asin Bsin C$, where $R$ is the circumradius of the triangle. Why? Because $2Rsin A=a$ and $2Rsin B=b$, we can reduce to the more familiar area formula $frac12absin C$. Maximizing $sin Asin Bsin C$ is now equivalent to maximizing the area of an acute triangle inscribed in a fixed circle $Gamma$.



                    And to do that, let's loosen things a bit, and instead maximize the area of any triangle inscribed in $Gamma$. If we fix two points $A$ and $B$ on the circle, what should $C$ be to maximize the area? It should be as far away from the line $AB$ as possible, namely one of the endpoints of the diameter of $Gamma$ that is the perpendicular bisector of $AB$. Comparing the two endpoints, the farther one is the one with the acute angle (or, if they're both right angles, both are equally far), so the area maximum with $A$ and $B$ fixed has $a=b$ and angle $C$ acute. Repeat for the other two pairs of vertices, and the maximum must come when $a=b=c$ and the triangle is equilateral. That's $60^circ$ angles $A,B,C$, for $x=y=z=frac{pi}{6}$ and $cos xcos ycos z = sin Asin Bsin C=left(frac{sqrt{3}}{2}right)^3=frac{3sqrt{3}}{8}$.



                    Now, that's a little unsatisfying logically - we proved that the maximum, if it exists, must be at an equilateral triangle. To patch this, we prove that there is a maximum. The area is a continuous function from $Gamma^3$ (the positions of the three points) to $mathbb{R}$. That's a compact domain, so the extreme value theorem applies and there is a maximum. Done.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Feb 2 at 6:34









                    jmerryjmerry

                    17.1k11633




                    17.1k11633












                    • $begingroup$
                      Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
                      $endgroup$
                      – KiWiJuice
                      Mar 14 at 7:44










                    • $begingroup$
                      While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
                      $endgroup$
                      – jmerry
                      Mar 14 at 7:48


















                    • $begingroup$
                      Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
                      $endgroup$
                      – KiWiJuice
                      Mar 14 at 7:44










                    • $begingroup$
                      While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
                      $endgroup$
                      – jmerry
                      Mar 14 at 7:48
















                    $begingroup$
                    Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
                    $endgroup$
                    – KiWiJuice
                    Mar 14 at 7:44




                    $begingroup$
                    Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
                    $endgroup$
                    – KiWiJuice
                    Mar 14 at 7:44












                    $begingroup$
                    While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
                    $endgroup$
                    – jmerry
                    Mar 14 at 7:48




                    $begingroup$
                    While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
                    $endgroup$
                    – jmerry
                    Mar 14 at 7:48


















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