How to solve this inequality involving trig functions: $\frac{\sin B + \sin C}{2}\leq \sin\left(\frac{B+C}{2}\right)$?

Question

Source - Larsen 1.3.12

Problem:

Prove using $y=\sin(x)$ graph, for $A,B,C$ as angles of triangle, that:

$$\frac{\sin B + \sin C}{2}\leq \sin\left(\frac{B+C}{2}\right)$$

Attempt:

(without graph)

Using sum-to-product formula, $\sin B+\sin C=2\sin(\frac{B+C}{2})\cos(\frac{B-C}{2})$

So,

$$\frac{\sin B + \sin C}{2}=\sin(\frac{B+C}{2})\cos(\frac{B-C}{2})$$ $$=\sin(\frac{180-A}{2})\cos(\frac{B-C}{2})$$ $$=\cos(\frac{A}{2})\cos(\frac{B-C}{2})$$ $$=\frac{\cos(\frac{A+B-C}{2})+\cos(\frac{A-B+C}{2})}{2}$$ $$=\frac{\cos(\frac{180-2C}{2})+\cos(\frac{180-2B}{2})}{2}$$ $$=\frac{\sin C+\sin B}{2}$$

And I think I arrived where I had begun.

Question:

How to solve this inequality? (without graph)

Answer 1

Based on Ekaveera's hint, right at the first step we have:

$$\frac{\sin B + \sin C}{2}=\sin(\frac{B+C}{2})\cos(\frac{B-C}{2})$$

Now, since $-1<\cos A<1$ for any $0<A<180$, we have:

$$\frac{\sin B + \sin C}{2} \leq \sin(\frac{B+C}{2})$$

Answer 2

Another proof, based on properties of the sine function is that it is true because, on $[0,\pi]$, sine is a concave function, since $\;(\sin)''(x)=-\sin x\le 0$ on this interval.