<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] Because $$\sum_{cyc}\left(\frac{a^2+bc}{b+c}-a\right)=\frac{\sum\limits_{cyc}(a^4-a^2b^2)}{\prod\limits_{cyc}(a+b)}=\frac{\sum\limits_{cyc}(a^2-b^2)^2}{2\prod\limits_{cyc}(a+b)}\geq0.$$
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