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

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: $(x;y;z) \in [(1;1;1);(2;2;0);(0;2;2);(2;0;2)]$
