<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] Answer:
Multiplying a in the numerator and the denominator, you get
$$LHS = \frac{\sum a^{2}}{\sum (ac+b^{2}a)}$$
Applying the Cauchy-Schwarz inequality  in the below steps
\begin{align}
&\geqslant\frac{(\sum a)^{2}}{\sum ac + \sum b^{2}a}\\
&\geqslant \frac{1}{\sum ac + \frac{1}{3}\sum a \sum a^2} = \frac{1}{3\sum ac +\sum a (\sum a)^2}\\
&\geqslant \frac{3}{\sum ac + (\sum a)^{2}}
\end{align}
Applying the inequality $$\sum ac \leqslant \frac{1}{3} (\sum a)^{2}$$
$$LHS \geqslant \frac{9}{4}$$
Edit: @9rm, Apply the Cauchy-Schwarz inequlality to $$\sum b^{2}a \leqslant (\sum b^4)^\frac{1}{2}\cdot (\sum a^2)^\frac{1}{2}.$$ This will reduce to $$\leqslant (abc)^\frac{1}{2}\cdot \sum a\cdot \sqrt{\frac{1}{3}}({\sum a})^{2}.$$ This will further reduce to $$\sqrt{\frac{1}{3}}\cdot \sum a\cdot \sqrt{\frac{1}{3}}\cdot {\sum a}^{2} = \frac{1}{3}\cdot \sum a(\sum a)^{2}.$$
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