Given a triangle with coners $A,B,C$, Show $sin(A)+sin(B)+sin(C) leq frac{3sqrt{3}}{2}$ [duplicate]











up vote
0
down vote

favorite
1













This question already has an answer here:




  • Maximum value of $sin A+sin B+sin C$?

    5 answers




Given a triangle with vertices $A,B,C$, Show $sin(A)+sin(B)+sin(C) leq frac{3 sqrt{3}}{2}$.



Here is a proof using Jensen inequality: $sin(x)$ is concave from $0$ to $pi$.
hence $frac{sin(A)+sin(B)+sin(C)}{3} leq sin(frac{A+B+C}{3})=frac{sqrt 3}{2}$ and hence we get the desired inequality by multiplying the previous inequality by $3$.










share|cite|improve this question















marked as duplicate by lab bhattacharjee algebra-precalculus
Users with the  algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 17 at 2:36


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.



















    up vote
    0
    down vote

    favorite
    1













    This question already has an answer here:




    • Maximum value of $sin A+sin B+sin C$?

      5 answers




    Given a triangle with vertices $A,B,C$, Show $sin(A)+sin(B)+sin(C) leq frac{3 sqrt{3}}{2}$.



    Here is a proof using Jensen inequality: $sin(x)$ is concave from $0$ to $pi$.
    hence $frac{sin(A)+sin(B)+sin(C)}{3} leq sin(frac{A+B+C}{3})=frac{sqrt 3}{2}$ and hence we get the desired inequality by multiplying the previous inequality by $3$.










    share|cite|improve this question















    marked as duplicate by lab bhattacharjee algebra-precalculus
    Users with the  algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.

    StackExchange.ready(function() {
    if (StackExchange.options.isMobile) return;

    $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
    var $hover = $(this).addClass('hover-bound'),
    $msg = $hover.siblings('.dupe-hammer-message');

    $hover.hover(
    function() {
    $hover.showInfoMessage('', {
    messageElement: $msg.clone().show(),
    transient: false,
    position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
    dismissable: false,
    relativeToBody: true
    });
    },
    function() {
    StackExchange.helpers.removeMessages();
    }
    );
    });
    });
    Nov 17 at 2:36


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

















      up vote
      0
      down vote

      favorite
      1









      up vote
      0
      down vote

      favorite
      1






      1






      This question already has an answer here:




      • Maximum value of $sin A+sin B+sin C$?

        5 answers




      Given a triangle with vertices $A,B,C$, Show $sin(A)+sin(B)+sin(C) leq frac{3 sqrt{3}}{2}$.



      Here is a proof using Jensen inequality: $sin(x)$ is concave from $0$ to $pi$.
      hence $frac{sin(A)+sin(B)+sin(C)}{3} leq sin(frac{A+B+C}{3})=frac{sqrt 3}{2}$ and hence we get the desired inequality by multiplying the previous inequality by $3$.










      share|cite|improve this question
















      This question already has an answer here:




      • Maximum value of $sin A+sin B+sin C$?

        5 answers




      Given a triangle with vertices $A,B,C$, Show $sin(A)+sin(B)+sin(C) leq frac{3 sqrt{3}}{2}$.



      Here is a proof using Jensen inequality: $sin(x)$ is concave from $0$ to $pi$.
      hence $frac{sin(A)+sin(B)+sin(C)}{3} leq sin(frac{A+B+C}{3})=frac{sqrt 3}{2}$ and hence we get the desired inequality by multiplying the previous inequality by $3$.





      This question already has an answer here:




      • Maximum value of $sin A+sin B+sin C$?

        5 answers








      algebra-precalculus trigonometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 16 at 0:52









      achille hui

      94.3k5129252




      94.3k5129252










      asked Nov 15 at 23:48









      mathnoob

      1,133115




      1,133115




      marked as duplicate by lab bhattacharjee algebra-precalculus
      Users with the  algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.

      StackExchange.ready(function() {
      if (StackExchange.options.isMobile) return;

      $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
      var $hover = $(this).addClass('hover-bound'),
      $msg = $hover.siblings('.dupe-hammer-message');

      $hover.hover(
      function() {
      $hover.showInfoMessage('', {
      messageElement: $msg.clone().show(),
      transient: false,
      position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
      dismissable: false,
      relativeToBody: true
      });
      },
      function() {
      StackExchange.helpers.removeMessages();
      }
      );
      });
      });
      Nov 17 at 2:36


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by lab bhattacharjee algebra-precalculus
      Users with the  algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.

      StackExchange.ready(function() {
      if (StackExchange.options.isMobile) return;

      $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
      var $hover = $(this).addClass('hover-bound'),
      $msg = $hover.siblings('.dupe-hammer-message');

      $hover.hover(
      function() {
      $hover.showInfoMessage('', {
      messageElement: $msg.clone().show(),
      transient: false,
      position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
      dismissable: false,
      relativeToBody: true
      });
      },
      function() {
      StackExchange.helpers.removeMessages();
      }
      );
      });
      });
      Nov 17 at 2:36


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























          2 Answers
          2






          active

          oldest

          votes

















          up vote
          2
          down vote













          You want maximize $$ f(A,B,C)=sin A+sin B+sin C$$ subject to $$ A+B+C=180$$



          Lagrange multipliers implies $$ (cos A,cos B ,cos C)=(1,1,1)$$ which gives you $$A=B=C=60$$



          That gives the maximum value of $3sqrt 3/2$






          share|cite|improve this answer




























            up vote
            1
            down vote













            This is a proof without Jensen's inequality. It mimic the proof for AM $ge$ GM for $3$ items.



            Notice for any $x, y in (0,pi)$, $$sin x + sin y = 2sinfrac{x+y}{2}cosfrac{x-y}{2} le 2sinfrac{x+y}{2}$$



            So for any $x,y,z in (0,pi)$, if we define $w = frac{x+y+z}{3}$, we will have



            $$begin{align}sin x + sin y + sin z + sin w
            &le 2sinfrac{x+y}{2} + 2sinfrac{z+w}{2}\
            & le 4sinfrac{x+y+z+w}{4} \
            &= 4sin w
            end{align}$$

            This leads to
            $$sin x + sin y + sin z le 3 sin w = 3 sin frac{x+y+z}{3}$$



            For any non-degenerate triangle, its angle $A,B,C in (0,pi)$ and $A+B+C = pi$. Above result implies



            $$sin A + sin B + sin C le 3 sinfrac{pi}{3} = frac{3sqrt{3}}{2}$$






            share|cite|improve this answer




























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              2
              down vote













              You want maximize $$ f(A,B,C)=sin A+sin B+sin C$$ subject to $$ A+B+C=180$$



              Lagrange multipliers implies $$ (cos A,cos B ,cos C)=(1,1,1)$$ which gives you $$A=B=C=60$$



              That gives the maximum value of $3sqrt 3/2$






              share|cite|improve this answer

























                up vote
                2
                down vote













                You want maximize $$ f(A,B,C)=sin A+sin B+sin C$$ subject to $$ A+B+C=180$$



                Lagrange multipliers implies $$ (cos A,cos B ,cos C)=(1,1,1)$$ which gives you $$A=B=C=60$$



                That gives the maximum value of $3sqrt 3/2$






                share|cite|improve this answer























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  You want maximize $$ f(A,B,C)=sin A+sin B+sin C$$ subject to $$ A+B+C=180$$



                  Lagrange multipliers implies $$ (cos A,cos B ,cos C)=(1,1,1)$$ which gives you $$A=B=C=60$$



                  That gives the maximum value of $3sqrt 3/2$






                  share|cite|improve this answer












                  You want maximize $$ f(A,B,C)=sin A+sin B+sin C$$ subject to $$ A+B+C=180$$



                  Lagrange multipliers implies $$ (cos A,cos B ,cos C)=(1,1,1)$$ which gives you $$A=B=C=60$$



                  That gives the maximum value of $3sqrt 3/2$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 16 at 0:06









                  Mohammad Riazi-Kermani

                  40.3k41958




                  40.3k41958






















                      up vote
                      1
                      down vote













                      This is a proof without Jensen's inequality. It mimic the proof for AM $ge$ GM for $3$ items.



                      Notice for any $x, y in (0,pi)$, $$sin x + sin y = 2sinfrac{x+y}{2}cosfrac{x-y}{2} le 2sinfrac{x+y}{2}$$



                      So for any $x,y,z in (0,pi)$, if we define $w = frac{x+y+z}{3}$, we will have



                      $$begin{align}sin x + sin y + sin z + sin w
                      &le 2sinfrac{x+y}{2} + 2sinfrac{z+w}{2}\
                      & le 4sinfrac{x+y+z+w}{4} \
                      &= 4sin w
                      end{align}$$

                      This leads to
                      $$sin x + sin y + sin z le 3 sin w = 3 sin frac{x+y+z}{3}$$



                      For any non-degenerate triangle, its angle $A,B,C in (0,pi)$ and $A+B+C = pi$. Above result implies



                      $$sin A + sin B + sin C le 3 sinfrac{pi}{3} = frac{3sqrt{3}}{2}$$






                      share|cite|improve this answer

























                        up vote
                        1
                        down vote













                        This is a proof without Jensen's inequality. It mimic the proof for AM $ge$ GM for $3$ items.



                        Notice for any $x, y in (0,pi)$, $$sin x + sin y = 2sinfrac{x+y}{2}cosfrac{x-y}{2} le 2sinfrac{x+y}{2}$$



                        So for any $x,y,z in (0,pi)$, if we define $w = frac{x+y+z}{3}$, we will have



                        $$begin{align}sin x + sin y + sin z + sin w
                        &le 2sinfrac{x+y}{2} + 2sinfrac{z+w}{2}\
                        & le 4sinfrac{x+y+z+w}{4} \
                        &= 4sin w
                        end{align}$$

                        This leads to
                        $$sin x + sin y + sin z le 3 sin w = 3 sin frac{x+y+z}{3}$$



                        For any non-degenerate triangle, its angle $A,B,C in (0,pi)$ and $A+B+C = pi$. Above result implies



                        $$sin A + sin B + sin C le 3 sinfrac{pi}{3} = frac{3sqrt{3}}{2}$$






                        share|cite|improve this answer























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          This is a proof without Jensen's inequality. It mimic the proof for AM $ge$ GM for $3$ items.



                          Notice for any $x, y in (0,pi)$, $$sin x + sin y = 2sinfrac{x+y}{2}cosfrac{x-y}{2} le 2sinfrac{x+y}{2}$$



                          So for any $x,y,z in (0,pi)$, if we define $w = frac{x+y+z}{3}$, we will have



                          $$begin{align}sin x + sin y + sin z + sin w
                          &le 2sinfrac{x+y}{2} + 2sinfrac{z+w}{2}\
                          & le 4sinfrac{x+y+z+w}{4} \
                          &= 4sin w
                          end{align}$$

                          This leads to
                          $$sin x + sin y + sin z le 3 sin w = 3 sin frac{x+y+z}{3}$$



                          For any non-degenerate triangle, its angle $A,B,C in (0,pi)$ and $A+B+C = pi$. Above result implies



                          $$sin A + sin B + sin C le 3 sinfrac{pi}{3} = frac{3sqrt{3}}{2}$$






                          share|cite|improve this answer












                          This is a proof without Jensen's inequality. It mimic the proof for AM $ge$ GM for $3$ items.



                          Notice for any $x, y in (0,pi)$, $$sin x + sin y = 2sinfrac{x+y}{2}cosfrac{x-y}{2} le 2sinfrac{x+y}{2}$$



                          So for any $x,y,z in (0,pi)$, if we define $w = frac{x+y+z}{3}$, we will have



                          $$begin{align}sin x + sin y + sin z + sin w
                          &le 2sinfrac{x+y}{2} + 2sinfrac{z+w}{2}\
                          & le 4sinfrac{x+y+z+w}{4} \
                          &= 4sin w
                          end{align}$$

                          This leads to
                          $$sin x + sin y + sin z le 3 sin w = 3 sin frac{x+y+z}{3}$$



                          For any non-degenerate triangle, its angle $A,B,C in (0,pi)$ and $A+B+C = pi$. Above result implies



                          $$sin A + sin B + sin C le 3 sinfrac{pi}{3} = frac{3sqrt{3}}{2}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 16 at 0:28









                          achille hui

                          94.3k5129252




                          94.3k5129252















                              Popular posts from this blog

                              Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

                              ComboBox Display Member on multiple fields

                              Is it possible to collect Nectar points via Trainline?