<s>[INST] 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
 [/INST] 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-Schwarz 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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