<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.
Since $(x+y+z)^2\geq3(xy+xz+yz),$ by C-S we obtain:
$$\sum_{cyc}\frac{a^2+bc}{b+c}=\sqrt{\left(\sum_{cyc}\frac{a^2+bc}{b+c}\right)^2}\geq\sqrt{3\sum_{cyc}\frac{(a^2+bc)(b^2+ac)}{(b+c)(a+c)}}=$$
$$=\sqrt{3(a^2+b^2+c^2)}\geq a+b+c.$$
The following inequality is also true.

Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq3\left(\frac{a^{\frac{20}{9}}+b^{\frac{20}{9}}+c^{\frac{20}{9}}}{3}\right)^{\frac{9}{20}}.$$

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