Prove that $\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}$ 
My question:
Let $a,b,c$ be positive real numbers satisfy $a+b+c=3.$
Prove that $$\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}.$$

I have tried to change the LHS to $$\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb^2}$$
And using Cauchy–Schwarz inequality for it
$$\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb^2}\geq \frac{(a+b+c)^2}{ab+bc+ca+ac^2+ba^2+cb^2}$$
Then because $$ab+ca+ca\leq \frac{(a+b+c)^2}{3}=\frac{3^2}{3}=3,$$
$$\frac{(a+b+c)^2}{ab+bc+ca+ac^2+ba^2+cb^2}\geq \frac{9}{3+ac^2+ba^2+cb^2}$$
Finally, I can't prove $ac^2+ba^2+cb^2\leq 3$ $$ $$
I look forward to your help, thank you!

Remarks: My proof of (1) is not nice. Hope to see a nice proof of it.
My proof:
Since the inequality is cyclic, WLOG, assume that $c = \min(a, b, c)$.
We split into two cases:
Case 1: $c \ge 1/5$
Using Cauchy-Bunyakovsky-Schwarz inequality, we have
$$\mathrm{LHS} \ge \frac{(a + b + c)^2}{a(b + c^2) + b(c + a^2) + c(a + b^2)}.$$
It suffices to prove that
$$ab + bc + ca + a^2b + b^2c + c^2 a \le 6 \tag{1}$$
which is true (the proof is given at the end).
Case 2: $c < 1/5$
Using $c^2 \le c < 1/5$, we have
$$\mathrm{LHS} \ge \frac{a}{b + c} + \frac{b}{1/5 + a^2} \ge \frac{a}{3 - a} + \frac{3 - a - 1/5}{1/5 + a^2}.$$
It suffices to prove that
$$\frac{a}{3 - a} + \frac{3 - a - 1/5}{1/5 + a^2} \ge \frac32$$
or (after clearing the denominators)
$$25a^3 - 35a^2 - 53a + 75 \ge 0$$
which is true (actually for all $a\ge 0$).
(Hint: Using AM-GM, we have $a^3 + \frac94 a \ge 3a^2$. The rest is easy.)
We are done.

Proof of (1):

We split into two cases:


(1) If $a < 1$ or $b < 1$, using the well-known inequality $a^2b + b^2c + c^2a + abc \le \frac{4}{27}(a + b + c)^3$, it suffices to prove that $$ab + bc + ca + 4 - abc \le 6$$ which is written as $$(1 - a)(b - 1)(b + a - 2) + (ab - a - b)(a + b + c - 3) \ge 0$$ which is true (using $a + b > 2$).


(2) If $a, b \ge 1$, let $a = 1 + u$ and $b = 1 + v$ for $u, v \ge 0$, and the inequality is written as $$-u^3 - 3u^2v + v^3 + u^2 + uv + v^2 \ge 0.$$ From $c = 3 - a - b = 1 - u - v$ and $c \ge 1/5$, we have $u + v \le 4/5$. Using $-u^3 \ge - u^2 \cdot \frac45$, it suffices to prove that $$-u^2\cdot \frac45 - 3u^2v + v^3 + u^2 + uv + v^2 \ge 0$$ or $$(1/5 - 3v)u^2 + v^3 + uv + v^2 \ge 0.$$


If $1/5 -3v \ge 0$, the inequality is true.


If $1/5 -3v < 0$, let $g(u) := (1/5 - 3v)u^2 + v^3 + uv + v^2$. Then $g(u)$ is concave. Note that $0 \le u \le 4/5 - v$. Also, we have $g(0) \ge 0$ and $g(4/5 - v) = \frac{-250v^3 + 625v^2 - 180v + 16}{125} \ge 0$. Thus, $g(u) \ge 0$ on $[0, 4/5 - v]$. The desired result follows.


We are done.

