How prove $\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\ge 1$ 
let $a,b,c$ be postive real numbers ,and such $$ab+bc+ac=3$$
  show that
  $$\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge 1$$

This problem is from Secrets In Inequalities volume 1  page 30,example 1.24. the comment.the  author this case is a bit more diffcult,But this author can't post solution.Thank you


By AM-GM (with all sums being cyclic), 
$$\sum \frac{a^2}{a+2b^2} = \sum a - \sum \frac{2ab^2}{a+2b^2}\ge \sum a - \sum \frac{2ab^2}{3\sqrt[3]{ab^4}}= \sum a - \frac23\sum (ab)^{2/3}$$
By Power Mean Inequality,  $$1 = \frac{ab+bc+ca}3 \ge \sqrt[\frac23]{\frac{\sum (ab)^{2/3}}3} \implies \sum (ab)^{2/3} \le 3$$
It remains to show that $\sum a \ge 3$ which follows from the well known
$$\left(\sum a\right)^2 \ge 3\sum ab = 9$$
