<s>[INST] Prove $5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+10.$ 
Problem. (?) For $a,b,c$ be non-negative numbers such as $a \geq 2(b+c).$
Prove:$$5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq
\frac{a^2+b^2+c^2}{ab+bc+ca}+10.$$

My Solution.
We write the inequality as
$$\big[2(b+c)-a\big]^2 \big[4\left( {{\mkern 1mu} b + {\mkern 1mu} c} \right){a^2} + 5{\mkern 1mu} \left( {2{\mkern 1mu} c + b} \right)\left( {c + 2{\mkern 1mu} b} \right)a + 3{\mkern 1mu} \left( {b + c} \right)\left( {6{\mkern 1mu} {b^2} + 19{\mkern 1mu} bc + 6{\mkern 1mu} {c^2}} \right)\big]$$
$$+\big[a-2(b+c)\big]\left( {b}^{2}+3bc+{c}^{2} \right)  \left( 36{b}^{2}+77bc+36{ c}^{2} \right)+2bc \left( b+c \right)  \left( b-c \right) ^{2}\geq 0,$$
which is true.
You can see the text to check here. I'm hoping for alternative proof..
Thank you!
 [/INST] Let $c=\min\{a,b,c\}$, $b=c+u$ and $a-2(b+c)=v$.
Thus, $u\geq0$, $v\geq0$, $a=4c+2u+v$ and
$$(ab+ac+bc)\prod_{cyc}(a+b)\left(5\sum_{cyc}\frac{a}{b+c}-\frac{a^2+b^2+c^2}{ab+ac+bc}-10\right)=$$
$$=745vc^4+2(2u^2+745uv+247v^2)c^3+(6u^3+1074u^2v+741uv^2+109v^3)c^2+$$
$$+(2u^4+329u^3v+355u^2v^2+109uv^3+8v^4)c+$$
$$+2uv(18u^3+27u^2v+13uv^2+2v^3)\geq0.$$
</s>