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.
 A: $$\sin A+ \sin B + \sin C=\sin A+ \sin B+ \sin (A+B)$$ $$= \sin(A)(1+ \cos (B))+ \sin(B)(1+\cos (A)) = \frac1{\sqrt3}((\sqrt3 \sin(A))(1+\cos(B))+(\sqrt3 \sin(B))(1+\cos(A))$$ 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 $a=b$.
A: 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 $$\frac{f(A)+f(B)+f(C)}{3}\leq f\Big(\frac{A+B+C}{3}\Big)\\\text{i.e.}\space\frac{\sin A+\sin B+\sin C}{3}\leq \sin \Big(\frac{A+B+C}{3}\Big)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\\\text{i.e.}\space \sin A+\sin B+\sin C\leq \frac{3\sqrt{3}}{2}\space\space\space \blacksquare$$
A: Let $a$, $b$ and $c$ be sides-lengths of the triangle and $S$ be an area of the triangle.
We need to prove that
$$\sum_{cyc}\frac{2S}{bc}\leq\frac{3\sqrt3}{2}$$ or
$$4S(a+b+c)\leq3\sqrt3abc$$ or
$$(a+b+c)^3(a+b-c)(a+c-b)(b+c-a)\leq27a^2b^2c^2.$$
Now, let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$. 
Hence, $x$, $y$ and $z$ are positives and we need to prove that
$$(x+y+z)^3xyz\leq\frac{27}{64}(x+y)^2(x+z)^2(y+z)^2.$$
Now, $$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)\Leftrightarrow$$
$$\Leftrightarrow\sum_{cyc}(x^2y+x^2z-2xyz)\geq0,$$
which is true by AM-GM.
Indeed, $$\sum_{cyc}(x^2y+x^2z)=\sum_{cyc}(y^2z+x^2z)\geq\sum_{cyc}(2\sqrt{y^2z\cdot x^2z})=2\sum_{cyc}xyz.$$
Thus, it remains to prove that
$$(xy+xz+yz)^2\geq3xyz(x+y+z),$$
which is AM-GM again. 
Indeed, $$(xy+xz+yz)^2-3xyz(x+y+z)=\sum_{cyc}(x^2y^2-x^2yz)=\frac{1}{2}\sum_{cyc}(2x^2y^2-2x^2yz)=$$
$$=\frac{1}{2}\sum_{cyc}(x^2y^2+x^2z^2-2x^2yz)\geq\frac{1}{2}\sum_{cyc}\left(2\sqrt{x^2y^2\cdot x^2z^2}-2x^2yz\right)=0.$$
Done!
