<s>[INST] Find the maxium $\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}$ 1) $a, b, c$ are triangle edges's length such that $abc = 1$. Find max:
$$\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}$$
My idea: $$\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}=\frac{abc}{a^3(b+c)}+\frac{abc}{b^{3}(c+a)}+\frac{abc}{c^3(a+b)}$$
Then use AM-GM ? I just can find min: $$\frac{1}{a^{3}\left ( b+c \right )}+\frac{1}{b^{3}\left ( c+a \right )}+\frac{1}{c^{3}\left ( a+b \right )}\geq \frac{3}{2}$$
2) Find for $x$, $y$, $z$ such that $\left\{\begin{matrix}
xy + 2(x+y)=0\\ 
\ yz + 2(y+z)=-3\\ 
zx + 2(z+x)=5
\end{matrix}\right.$
(Some one should edit my post: correct grammar...)
 [/INST] Hint: For your second question, subtracting the equations in pairs suggests the substitution $$a=x+2,\quad b=y+2, \quad c=z+2.$$
Try this in the original set of equations. 
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