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.

Another way.
$$\sum_{cyc}\frac{a^2+bc}{b+c}-a-b-c=\sum_{cyc}\left(\frac{a^2+bc}{b+c}-\frac{b+c}{2}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\frac{2a^2-b^2-c^2}{b+c}=\frac{1}{2}\sum_{cyc}\frac{a^2-b^2-(c^2-a^2)}{b+c}=$$
$$=\frac{1}{2}\sum_{cyc}(a^2-b^2)\left(\frac{1}{b+c}-\frac{1}{a+c}\right)=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2(a+b)}{(a+c)(b+c)}\geq0.$$
