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 : $$ \sin x +\sin y +\sin z\le \frac{3}{2} $$

With little effort, one can prove that $$\sin x = \frac{1}{\sqrt{1+(2\cot\alpha+\cot\beta)^2}}$$, where $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$. It's also easy to check that the cotangents of angles of any triangle satisfies the following: $$\cot\alpha\cot\beta+\cot\beta\cot\gamma+\cot\gamma\cot\alpha = 1$$. Therefore, the problem is equivalent to the following inequality: $$x,y,z\in\mathbb{R} : xy+yz+zx=1\rightarrow\sum_{x,y,z}\sqrt\frac{1}{1+(2x+y)^2}\leq\frac{3}{2}$$ Now, while I believe that an avid contest-type inequality solver could find an elegant solution to above, it can also be solved by Lagrange multiplier method. To begin with, let's make the following substituion $$(2x+y, 2y+z, 2z+x)= (a,b,c)\leftrightarrow (x,y,z) = (\frac{4a-2b+c}{9},\frac{4b-2c+a}{9},\frac{4c-2a+b}{9})$$ Now our problem is equivalent to $$-2(a^2+b^2+c^2)+5(ab+bc+ca) = 27\rightarrow \sum_{a,b,c}\frac{1}{\sqrt {1+a^2}}\leq\frac{3}{2}$$. Now it's a routine exercise of Lagrange multiplier method that yields the inequality is attained where $a=b=c=3x=3y=3z$, which means our triangle must be equilateral for this to hold.

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!