<s>[INST] Showing $\sum_{cyc}\frac{a^2+bc}{b+c}\geq a+b+c$ for positive $a$, $b$, $c$ The following is an inequality which I have trouble solving:
$$\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$$
($a, b, c>0$)
I tried multiplying LHS and RHS by 2 and then use
$${2(a+b+c)}={(a+b)+(b+c)+(c+a)}$$
and move to LHS but did not succeed in solving by this means.
Please provide a solution.
 [/INST] Another way.
By C-S $$\sum_{cyc}\frac{a^2+bc}{b+c}=\sum_{cyc}\frac{(a^2+bc)^2}{(a^2+bc)(b+c)}\geq\frac{\left(\sum\limits_{cyc}(a^2+bc)\right)^2}{\sum\limits_{cyc}(a^2+bc)(b+c)}=$$
$$=\frac{\sum\limits_{cyc}(a^4+2a^2b^2+a^2b^2+2a^2bc+2a^3b+2a^3c+2a^2bc)}{2\sum\limits_{cyc}(a^2b+a^2c)}=$$
$$=a+b+c+\frac{\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+4a^2bc)}{2\sum\limits_{cyc}(a^2b+a^2c)}-\sum_{cyc}a=$$
$$=a+b+c+\frac{\sum\limits_{cyc}(a^4-a^2b^2)}{2\sum\limits_{cyc}(a^2b+a^2c)}=a+b+c+\frac{\sum\limits_{cyc}(a^2-b^2)^2}{4\sum\limits_{cyc}(a^2b+a^2c)}\geq a+b+c.$$
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