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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...)

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two unrelated questions in one. why don't you separate them. –  user31280 Nov 23 '12 at 11:32
    
I think we can't find max of $$\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}$$ –  Xeing Nov 23 '12 at 12:47

2 Answers 2

up vote 2 down vote accepted

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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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 $I$.

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