If $a+b+c=1$ and a,b,c ＞0 prove $\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$

Question

If $a+b+c=1$ and a,b,c＞0 prove $\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$. I tried with CS Engel form,homogenization but ina anyway i can't prove inequality. Can someone helpp?

Answer 1

It's wrong for real numbers.

For positive variables by C-S $$\sum_{cyc}\frac{a^2}{a^2+c}=\sum_{cyc}\frac{a^2(a+b)^2}{(c(a+b+c)+a^2)(a+b)^2}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(a^2+c^2+ac+bc)(a+b)^2}.$$ Thus, it's enough to prove that $$4\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq3\sum\limits_{cyc}(a^2+c^2+ac+bc)(a+b)^2$$ or $$\sum_{cyc}(a^4-a^3b+5a^3c+3a^2b^2-8a^2bc)\geq0,$$ which is true because $$\sum_{cyc}(a^4-a^3b)\geq0$$ by Rearrangement; $$\sum_{cyc}a^3c\geq\sum_{cyc}a^2bc$$ it's $$\sum_{cyc}\frac{a^2}{b}\geq\sum_{cyc}a,$$ which is true by Rearrangement again and $$\sum_{cyc}(a^2b^2-a^2bc)=\frac{1}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$