Prove: $ \sum\limits_{cyc}\frac{ab}{a^2+b^2}+\frac{1}{4}(\sum\limits_{cyc}\frac{1}{a})\geq\frac{15}{4} $ Let $a,b,c>0$ such that $a+b+c=1$.
Prove: $ \sum\limits_{cyc}\frac{ab}{a^2+b^2}+\frac{1}{4}\sum\limits_{cyc}\frac{1}{a}\geq\frac{15}{4} $
I don't have any idea. You guy have any idea?? 

By AM-GM inequality:
$$\frac{ab}{a^2+b^2}= \frac12-\frac{(a-b)^2}{2(a^2+b^2)}\ge \frac12-\frac{(a-b)^2}{4ab}=1-\frac14\left(\frac{a}b+\frac{b}a\right)$$
Also
$$\frac1{4a}= \frac{a+b+c}{4a}=\frac14+\frac14\left(\frac{b}a+\frac{c}a \right)$$
Cyclically summing the above two and adding, we get the required inequality.
