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.

$$\sin A+ \sin B + \sin C=\sin A+ \sin B+ \sin (A+B)$$ $$= \sin(A)(1+ \cos (B))+ \sin(B)(1+\cos (A)) = \frac1{\sqrt3}((\sqrt3 \sin(A))(1+\cos(B))+(\sqrt3 \sin(B))(1+\cos(A))$$ Then use A.M.-G.M. inequality, $$\le \frac1{2\sqrt3}(3\sin^2(A)+1+\cos^2(B)+2\cos(B)+3\sin^2(B)+1+\cos^2(A)+2\cos(A))$$
$$=\frac1{2\sqrt3}(2\sin^2(A)+2\sin^2(B)+4+2\cos(B)+2\cos(A))=\frac1{\sqrt3}( 1-\cos^2(A)+1-\cos^2(B)+2+\cos(B)+\cos(A))=\frac1{\sqrt3}\left(1+\frac14-\left(\cos(A)-\frac12\right)^2+1+\frac14-\left(\cos(B)-\frac12\right)^2+2\right)\le\frac1{\sqrt3}\left(4+\frac12\right)=\frac9{2\sqrt3}=\frac{3\sqrt3}2$$
I multiplied and divided $\sin(A)$ by $\sqrt3$ because I know that the function maximizes when $A=B=\frac{\pi}3$ and so $1+\cos(B)=\frac32=\sqrt3\times \sin(\frac{\pi}3)=\sqrt3\times \sin(A)$, because $ab\le \frac{a^2+b^2}2$ and equality holds only when $a=b$.
