<s>[INST] If $a,b,c>0$ and $a+b+c=1$ prove inequality: $\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$ If $a,b,c>0$ and $a+b+c=1$ prove inequality: $$\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$$
 [/INST] By C-S
$$\sum_{cyc}\frac{a}{b+c^2}=\sum_{cyc}\frac{a^2}{ab+ac^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+a^2b)}.$$
Thus, it remains to prove that
$$4(a+b+c)^2\geq9(ab+ac+bc)+9(a^2b+b^2c+c^2a)$$ or
$$4(a+b+c)^3\geq9(ab+ac+bc)(a+b+c)+9(a^2b+b^2c+c^2a)$$ or
$$\sum_{cyc}(4a^3-6a^2b+3a^2c-abc)\geq0$$
and since by AM-GM $\sum\limits_{cyc}a^2c\geq3abc$, it remains to prove that
$$\sum_{cyc}(2a^3-3a^2b+a^2c)\geq0$$ or
$$\sum_{cyc}(2a^3-3a^2b+ab^2)\geq0$$ or
$$\sum_{cyc}a(a-b)(2a-b)\geq0$$ or
$$\sum_{cyc}\left(a(a-b)(2a-b)-\frac{1}{3}(a^3-b^3)\right)\geq0$$ or
$$\sum_{cyc}(a-b)^2(5a+b)\geq0.$$
Done!
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