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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Maximum value of $sin A+sin B+sin C$?
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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 $3$.
algebra-precalculus trigonometry
edited Nov 16 at 0:52
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Maximum value of $sin A+sin B+sin C$?
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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 $3$.
algebra-precalculus trigonometry
edited Nov 16 at 0:52
achille hui
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asked Nov 15 at 23:48
mathnoob
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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 $3$.
algebra-precalculus trigonometry
edited Nov 16 at 0:52
achille hui
93.4k5127251
asked Nov 15 at 23:48
mathnoob
63511
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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 $3$.
This question already has an answer here:
Maximum value of $sin A+sin B+sin C$?
5 answers
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Nov 16 at 0:52
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asked Nov 15 at 23:48
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edited Nov 16 at 0:52
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edited Nov 16 at 0:52
achille hui
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edited Nov 16 at 0:52
achille hui
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edited Nov 16 at 0:52
achille hui
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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 $3sqrt 3/2$
answered Nov 16 at 0:06
Mohammad Riazi-Kermani
40.2k41958
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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
$$sin A + sin B + sin C le 3 sinfrac{pi}{3} = frac{3sqrt{3}}{2}$$
answered Nov 16 at 0:28
achille hui
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2 Answers
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2 Answers
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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 $3sqrt 3/2$
answered Nov 16 at 0:06
Mohammad Riazi-Kermani
40.2k41958
add a comment |
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 $3sqrt 3/2$
answered Nov 16 at 0:06
Mohammad Riazi-Kermani
40.2k41958
add a comment |
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 $3sqrt 3/2$
answered Nov 16 at 0:06
Mohammad Riazi-Kermani
40.2k41958
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 $3sqrt 3/2$
answered Nov 16 at 0:06
Mohammad Riazi-Kermani
40.2k41958
answered Nov 16 at 0:06
Mohammad Riazi-Kermani
40.2k41958
answered Nov 16 at 0:06
Mohammad Riazi-Kermani
40.2k41958
answered Nov 16 at 0:06
Mohammad Riazi-Kermani
40.2k41958
40.2k41958
add a comment |
add a comment |
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
$$sin A + sin B + sin C le 3 sinfrac{pi}{3} = frac{3sqrt{3}}{2}$$
answered Nov 16 at 0:28
achille hui
93.4k5127251
add a comment |
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
$$sin A + sin B + sin C le 3 sinfrac{pi}{3} = frac{3sqrt{3}}{2}$$
answered Nov 16 at 0:28
achille hui
93.4k5127251
add a comment |
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
$$sin A + sin B + sin C le 3 sinfrac{pi}{3} = frac{3sqrt{3}}{2}$$
answered Nov 16 at 0:28
achille hui
93.4k5127251
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
$$sin A + sin B + sin C le 3 sinfrac{pi}{3} = frac{3sqrt{3}}{2}$$
answered Nov 16 at 0:28
achille hui
93.4k5127251
answered Nov 16 at 0:28
achille hui
93.4k5127251
answered Nov 16 at 0:28
achille hui
93.4k5127251
answered Nov 16 at 0:28
achille hui
93.4k5127251
93.4k5127251
add a comment |
add a comment |
