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

Question

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 $3$.

Answer 1

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 $3\sqrt 3/2$

Answer 2

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

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

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

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

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