<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] Suppose $a+b+c=1,$ let $p=a+b+c,\,q=ab+bc+ca,\,r=abc$ then $p \leqslant \frac{1}{3}.$ We write inequality as
$$\frac{5(1-2q+3r)}{q-r} \geqslant \frac{1-2q}{q}+10,$$
or
$$(1+23q)r+2q(2-9q) \geqslant 0. \quad (1)$$
If $0 < q \leqslant \frac{2}{9}$ then $(1)$ is true.
If $\frac{2}{9} < q \leqslant \frac{1}{4},$ from condition we get
$$(a-2b-2c)(b-2c-2a)(c-2a-2b) \geqslant 0,$$
equivalent to
$$r \geqslant \frac{18q-4}{27}.$$
Therefore, we need to prove
$$(1+23q) \cdot \frac{18q-4}{27}+2q(2-9q) \geqslant 0,$$
or
$$\frac{2(9q-2)(1-4q)}{27} \geqslant 0 \quad (\text{true}).$$
If $\frac{1}{4} < q \leqslant \frac 13,$ then
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+10 \leqslant 12.$$
We need to prove
$$5\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right) \geqslant 12.$$
Let $t = \frac{a}{b+c} \geqslant 2,$ using the Cauchy-Schwarz inequality we have
$$\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{(b+c)^2}{a(b+c)+2bc} \geqslant \frac{2}{2t+1}.$$
It's remain to prove that
$$5t + \frac{10}{2t+1} \geqslant 12,$$
or
$$\frac{(10t+1)(t-2)}{2t+1} \geqslant 0.$$
Which is true.
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