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...)