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