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











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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 $3$.










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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.



















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    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 $3$.










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    marked as duplicate by lab bhattacharjee algebra-precalculus
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    Nov 17 at 2:36


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      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






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      edited Nov 16 at 0:52









      achille hui

      94.3k5129252




      94.3k5129252










      asked Nov 15 at 23:48









      mathnoob

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      1,133115




      marked as duplicate by lab bhattacharjee algebra-precalculus
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      Nov 17 at 2:36


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          2 Answers
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          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

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              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















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