<s>[INST] How to solve this inequality involving trig functions: $\frac{\sin B + \sin C}{2}\leq \sin\left(\frac{B+C}{2}\right)$? 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)
 [/INST] 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.
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