<s>[INST] Prove the inequality using AM-GM only. By considering "Arithmetic mean $\geq $ Geometric mean" prove the trigonometric inequality: 
$$\sin A + \sin B + \sin C \leq \frac{3\sqrt{3}}{2}. $$
where $A+B+C=180°$. 
My try: By using transformation formulae, I proved that
$$\sin A + \sin B + \sin C = 4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)=y(let)$$
Next using AM-GM inequality
$$\cos\left(\frac{A}{2}\right)+\cos\left(\frac{B}{2}\right)+\cos\left(\frac{C}{2}\right)\geq 3 \left(\frac{y}{4}\right)^{\frac{1}{3}}.$$
I'm unable to proceed further. Please help me.
 [/INST] You can use Jensen Inequality on concave function.
Let $f(x)=\sin x$, then $f''(x)=-\sin x<0$, as $x\in (0,\pi)$.
Then $$\frac{f(A)+f(B)+f(C)}{3}\leq f\Big(\frac{A+B+C}{3}\Big)\\\text{i.e.}\space\frac{\sin A+\sin B+\sin C}{3}\leq \sin \Big(\frac{A+B+C}{3}\Big)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\\\text{i.e.}\space \sin A+\sin B+\sin C\leq \frac{3\sqrt{3}}{2}\space\space\space \blacksquare$$
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