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: $$\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$$
 A: 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!
A: 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 $$\sqrt{\frac{1}{3}}\cdot \sum a\cdot \sqrt{\frac{1}{3}}\cdot {\sum a}^{2} = \frac{1}{3}\cdot \sum a(\sum a)^{2}.$$
A: 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!
