Rearrangement Inequality- $\sum_{c y c} \frac{a^{2}+b c}{b+c} \geq a+b+c$

Question

$$\sum_{c y c} \frac{a^{2}+b c}{b+c} \geq a+b+c$$

I'm confused about how to solve this. can someone give me a few hints? I'm stuck thing what even $\sum_{c y c}$ means!

Answer 1

It's just Power-Mean Inequality.
By Inequality of AM and HM, $$ \sum_{cyc}{a^2+bc\over b+c} \geq \sum_{cyc}{a} \Rightarrow \sum_{cyc}{\left({a^2+bc\over b}+{a^2+bc\over c}\right)} \geq 4\sum_{cyc}{a} $$ $$ \Rightarrow \sum_{cyc}{\left( {a^2\over b}+{b^2\over a} \right)} \geq 2\sum_{cyc}{a} $$ Notice that $$ {a^2\over b}+ b\ge 2a $$ and $$ {b^2\over a}+a\ge 2b $$ And by $4$ other similar inequalities summing up, we meet our desire.

Answer 2

$$ \sum_{cyc} \frac{a^2 + bc}{b+ c} = \frac{a^2 + bc}{b + c} + \frac{b^2 + ca }{c + a} + \frac{c^2 + ab}{a + b}$$ Is it clear now?