<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] since
$$\sum_{cyc}\dfrac{a}{b+c^2}=\sum_{cyc}\dfrac{a^2}{ab+ac^2}$$
Use Cauchy-Schwarz inequality,we have
$$\left(\sum_{cyc}\dfrac{a^2}{ab+ac^2}\right)\cdot\sum_{cyc}(ab+ac^2)\ge\left(\sum_{cyc}a\right)^2=1$$
so we only prove this following inequality
$$\dfrac{1}{\sum_{cyc}(ab+ac^2)}\ge\dfrac{9}{4}$$
$$\Longleftrightarrow 4\ge9\sum_{cyc}ab+9\sum_{cyc}ac^2$$
$$\Longleftrightarrow 4(\sum_{cyc}a)^2\ge 9\sum_{cyc}ab+9\sum_{cyc}ac^2$$
$$\Longleftrightarrow 4\sum_{cyc}a^2\ge \sum_{cyc}ab+9\sum_{cyc}ac^2$$
since
$$\sum_{cyc}a^2\ge \sum_{cyc}ab$$
so we only prove this
$$3\sum_{cyc}a^2\ge 9\sum_{cyc}ac^2$$
$$\Longleftrightarrow \sum_{cyc}a^2\ge3\sum_{cyc}ac^2$$
$$\Longleftrightarrow \sum_{cyc}a\cdot\sum_{cyc}a^2\ge3\sum_{cyc}ac^2$$
$$\Longleftrightarrow \sum_{cyc}a^3+\sum_{cyc}ab^2\ge 2\sum_{cyc}ac^2$$
since Use AM-GM inequality
$$\sum_{cyc}a^3+\sum_{cyc}ab^2=\sum_{cyc}(a^3+ab^2)\ge\sum_{cyc}(2a^2b)=2\sum_{cyc}ac^2$$
By done!
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