<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] 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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