Given:$$\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2} \le \frac{1}{4} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$$Homogenizing: $$\frac{a}{(a+b+c)a+b^2}+\frac{b}{(a+b+c)b+c^2}+\frac{c}{(a+b+c)c+a^2}=\frac{a}{a^2+ab+ac+b^2}+\frac{b}{ab+b^2+bc+c^2}+\frac{c}{ac+bc+c^2+a^2}$$AM-GM on denominators gives that we need to prove that $$\sum_{cyc}\frac{1}{\sqrt[4]{b^3c}}\le \sum_{cyc}\frac{1}{a},$$which is equivalent to $$\frac{a\sqrt[4]{bc^3}+b\sqrt[4]{a^3c}+c\sqrt[4]{ab^3}}{abc}\le \frac{ab+bc+ca}{abc},$$which is equivalent to $$ab^{\frac{1}{4}}c^{\frac{3}{4}}+bc^{\frac{1}{4}}a^{\frac{3}{4}}+ca^{\frac{1}{4}}b^{\frac{3}{4}}\le ab+bc+ca,$$which is true by weigthed AM-GM.

By AM-GM, $$\sum_{cyc}\frac{a}{a+b^2}=\sum_{cyc}\frac{a}{a(a+b+c)+b^2}\leq\sum_{cyc}\frac{1}{\sqrt[4]{b^3c}}\leq\frac{1}{16}\sum_{cyc}\left(\frac{3}{b}+\frac{1}{c}\right)=\frac{1}{4}\sum_{cyc}\frac{1}{a} .$$

parmenides51 wrote: Let $a, b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove that $$\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2} \le \frac{1}{4} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$$ Solution of Zhangyanzong: $$\sum_{cyc}\frac{a}{a+b^2}=\sum_{cyc}\frac{a}{a^2+b^2+ab+ca}\leq\sum_{cyc}\frac{1}{3b+c}\leq\frac{1}{16}\sum_{cyc}\left(\frac{3}{b}+\frac{1}{c}\right)=\frac{1}{4}\sum_{cyc}\frac{1}{a} $$

parmenides51 wrote: Let $a, b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove that $$\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2} \le \frac{1}{4} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$$ Solution: $$\iff$$$$ \frac{b^2}{a+b^2}+\frac{c^2}{b+c^2}+\frac{a^2}{c+a^2}+\frac{ab+bc+ca}{4abc}\geq 3.$$$$ab+bc+ca\leq\frac{(a+b+c)^2}{3}=\frac{1}{3}<\frac{3}{4}\implies $$$$(ab+bc+ca-\frac{1}{3})(ab+bc+ca-\frac{3}{4})\geq 0$$$$\iff $$$$\frac{1}{2-2(ab+bc+ca)}+\frac{3}{4(ab+bc+ca)}\geq 3$$$$\frac{b^2}{a+b^2}+\frac{c^2}{b+c^2}+\frac{a^2}{c+a^2}+\frac{ab+bc+ca}{4abc}$$$$\geq \frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}+\frac{ab+bc+ca}{4abc}\geq\frac{1}{2-2(ab+bc+ca)}+\frac{3}{4(ab+bc+ca)} \geq 3.$$

parmenides51 wrote: Let $a, b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove that $$\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2} \le \frac{1}{4} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$$ Solution of Zhangyanzong: $$\iff$$$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+4\left(\frac{b^2}{a+b^2}+\frac{c^2}{b+c^2}+\frac{a^2}{c+a^2}\right)\geq 12.$$$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+4\left(\frac{b^2}{a+b^2}+\frac{c^2}{b+c^2}+\frac{a^2}{c+a^2}\right)\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{4(a+b+c)^2}{a+b+c+a^2+b^2+c^2}$$$$\geq 4\sqrt[4]{\frac{4}{abc(1+a^2+b^2+c^2)}}\geq4\sqrt[4]{\frac{36}{(ab+bc+ca)(1+a^2+b^2+c^2)}}$$$$=4\sqrt[4]{\frac{72}{1-(a^2+b^2+c^2)^2}}\geq4\sqrt[4]{\frac{72}{1-\frac{1}{9}(a+b+c)^4}}=12.$$

sqing wrote: Let $a, b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+4\left(\frac{b^2}{a+b^2}+\frac{c^2}{b+c^2}+\frac{a^2}{c+a^2}\right)\geq 12.$$

Let $a, b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove that $$\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\geqslant \frac{13}{2(1-abc)}$$h