<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.
Let $a\geq b\geq c$.
Thus, $$\sum_{cyc}\left(\frac{a^2+bc}{b+c}-a\right)=\sum_{cyc}\frac{(a-b)(a-c)}{b+c}\geq\frac{(a-b)(a-c)}{b+c}+\frac{(b-a)(b-c)}{a+c}=$$
$$=(a-b)\left(\frac{a-c}{b+c}-\frac{b-c}{a+c}\right)\geq0$$ because $a-b\geq0,$ $a-c\geq b-c$ and $\frac{1}{b+c}\geq\frac{1}{a+c}.$
</s>