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
 A: 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!
A: A full expanding gives 
$$\sum_{cyc}(a^6+2a^5b+2a^5c-a^4b^2-a^4c^2-4a^3b^3+a^2b^2c^2)\geq0$$ or
$$\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)+2\sum_{cyc}ab(a^2-b^2)^2\geq0,$$
which is true by Schur.
Done!
A: 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: $(x;y;z) \in [(1;1;1);(2;2;0);(0;2;2);(2;0;2)]$
A: 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 $x=y=z=1$. 
