<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] For $0 < \epsilon < 1$ and $a = \epsilon^2$, $b = c = \frac 1 \epsilon$ we have
$$
abc = 1\\
a = \epsilon^2 \leq \frac 2 \epsilon = b + c \\
b = \frac 1 \epsilon \leq \frac 1 \epsilon + \epsilon^2 = a + c \\
c = \frac 1 \epsilon \leq \frac 1 \epsilon + \epsilon^2 = a + b
$$
All question conditions are satisfied.
Now
$$
I := \frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b} \geq
\frac{bc}{a^{2}b+a^{2}c} = \frac 1 {a^3(b +c)} = \frac 1 {2\epsilon^5}
$$
so for arbitrarily small $\epsilon$ we get arbitrarily great $I$.
</s>