inequality with three variables and condition

Question

If $a$,$b$ and $c$ positive real numbers such that $a+b+c=1$, prove $\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$. I have tried several methods to solve this,but can't get any result. Any idea?

Answer 1

By C-S and Vasc we obtain $\sum\limits_{cyc}\frac{a^2}{a^2+c}=\sum\limits_{cyc}\frac{a^4}{a^4+a^2c(a+b+c)}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+a^3c+a^2b^2+a^2bc)}\geq$

$\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+a^2b^2+a^2bc)+\frac{(a^2+b^2+c^2)^2}{3}}=\frac{3(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(4a^4+5a^2b^2+3a^2bc)}$.

Id est, it remains to prove that $4(a^2+b^2+c^2)^2\geq\sum\limits_{cyc}(4a^4+5a^2b^2+3a^2bc)$, which is $\sum\limits_{cyc}c^2(a-b)^2\geq0$. Done!