How to prove that $ \sin \angle{GAB}+\sin \angle{GBC}+\sin \angle{GCA} \le \frac{3}{2} $ for a triangle $ABC$ with centroid $G$? Let $ G $ be the centroid of $ \triangle ABC $ , such that  $ \measuredangle{GAB}=x,\measuredangle{GBC}=y,\measuredangle{GCA}=z $.
How do I prove  that :
$$  \sin x +\sin y +\sin z\le \frac{3}{2} $$

Let $AA'$, $BB'$ and $CC'$ be medians of the triangle.
Thus, from $\Delta ABA'$ we obtain:
$$\frac{\frac{a}{2}}{\sin{x}}=\frac{AA'}{\sin\beta}$$ or
$$\sin{x}=\frac{a\sin\beta}{2AA'}$$ or
$$\sin{x}=\frac{a\cdot\frac{2S}{ac}}{\sqrt{2b^2+2c^2-a^2}}$$ or
$$\sin{x}=\frac{2S}{c\sqrt{2b^2+2c^2-a^2}}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{1}{c\sqrt{2b^2+2c^2-a^2}}\leq\frac{3}{4S}$$ or
$$\sum_{cyc}\frac{1}{c\sqrt{2b^2+2c^2-a^2}}\leq\frac{3}{\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)}}.$$ 
Now, by C-S 
$$\left(\sum_{cyc}\frac{1}{c\sqrt{2b^2+2c^2-a^2}}\right)^2\leq\sum_{cyc}\frac{1}{c^2}\sum_{cyc}\frac{1}{2b^2+2c^2-a^2}.$$
Id est, it remains to prove that
$$\sum_{cyc}\frac{1}{a^2}\sum_{cyc}\frac{1}{2b^2+2c^2-a^2}\leq\frac{9}{\sum\limits_{cyc}(2a^2b^2-a^4)},$$
which is
$$(a^2-b^2)^2(a^2-c^2)^2(b^2-c^2)^2\geq0.$$
Done!
