# Showing that: $(\frac{a}{b+c})^2+(\frac{b}{a+c})^2+(\frac{c}{a+b})^2+\frac{10abc}{(a+b)(b+c)(c+a)}\ge 2$

Let a;b;c>0. Prove: $$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{a+c}\right)^2+\left(\frac{c}{a+b}\right)^2+\frac{10abc}{(a+b)(b+c)(c+a)}\geq 2$$

I think

$$\frac{2a}{b+c}=x;\frac{2b}{c+a}=y;\frac{2c}{a+b}=z$$

We have: $xy+yz+zx+xyz=4$

$$(\frac{x}{2})^2+(\frac{y}{2})^2+(\frac{z}{2})^2+\frac{10xyz}{8} \ge 2$$ $\Leftrightarrow x^2+y^2+z^2+5xyz \ge 8$ $\Leftrightarrow x^2+y^2+z^2-5(xy+yz+zx) +12 \ge 0$ deadlock

Can you help me? Thank you very much

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I collected a solution:

Need to prove: $x^2+y^2+z^2+5xyz \ge 8$

Put: $x+y+z=p;xy+yz+zx=q;xyz=r$

We have: $q+r=4$

Need to prove inequality is equivalent to:

$p^2-2q+5r \ge 8 \Leftrightarrow p^2-7q+12 \ge 0$

*)If: $p \le 4$. Applying Schur's inequality; we have:

$r \ge \dfrac{p(4q-p^2)}{9} \Rightarrow 4 \ge q+\dfrac{p(4q-p^2)}{9} \Leftrightarrow q \le \dfrac{p^3+36}{4p+9}$

WE will prove: $p^2-\dfrac{7(p^3+36)}{4p+9}+12 \ge 0 \Leftrightarrow (p-3)(p^2-16) \ge 0$ TRUE

Because: $4 \ge p \ge \sqrt{3q} \ge 3$

*) If: $p \ge 4$. We have:

$p^2-2q+5r \ge p^2-2q \ge \dfrac{p^2}{8} \ge 5$

Equality occurs if and only if: $(x;y;z) \in [(1;1;1);(2;2;0);(0;2;2);(2;0;2)]$

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Nice problem! Here is my solution.

Let $x=\frac{a}{b+c},y=\frac{b}{c+a},z=\frac{c}{a+b}$

Then we have to show that $$x^2+y^2+z^2+10xyz\geq 2$$ Fisrtly,we have the identity $$xy+yz+zx+2xyz=1$$ Secondly,we have the known inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}$$ that implys

$$x+y+z\geq \frac{3}{2}$$ Now Using Schur of the third degree,we have $$x^2+y^2+z^2+6xyz+4xyz\geq x^2+y^2+z^2+\frac{9xyz}{x+y+z}+4xyz\geq 2(xy+yz+xz)+4xyz=2$$ Hence we are done!

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Nice solution :) –  Sanchez Mar 9 '13 at 2:22

You are almost there. You have $x^{2}+y^{2}+z^{2}\ge xy+yz+xz$ by matching squares. So $$x^{2}+y^{2}+z^{2}-5(xy+yz+zx)+12\ge 12-4(xy+yz+zx)\ge 0$$ In other words you need to show $xy+yz+zx\le 3$. Now by Cauchy-Schwartz the left hand side only obtain maximum of $x=ky, y=kz,z=kx$. So we have either $x=y=z$ or $x=-y, y=-z,z=-x$. The second case is impossible since $a,b,c> 0$. Therefore the maximum is obtained when $x=y=z=1$.

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This is wrong. Use AM-GM or other tools, you can show that $xy+yz+zx \ge 3$. –  Sanchez Dec 23 '12 at 13:34
By making a change of variable to eliminate the cross terms, the left hand side becomes clearly indefinite: wolframalpha.com/input/… –  Hurkyl Dec 25 '12 at 1:31
I see where this got wrong. It seems difficult to tackle. –  user32240 Dec 25 '12 at 4:06