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!

Partial answer hint :
For $x,y,z\in[0.5,2]$   we have the inequality :
$$\frac{x}{y+z^{2}}-\frac{1}{x+y+z}\frac{x}{y+z}\ge0$$
Summing and using the constraint  gives the inequality .
Now we have one variable less than $1/2$ come back further .
Extended comment :
I have almost if finish if $\max({a,b,c})=a\ge 2$ the inequality is obvious . Next we have the following inequalities for $2\geq a\geq 1.5\geq c\ge 0.5\geq b$: $$\frac{a}{b+c^{2}}+\frac{c}{a+b^{2}}+\min\left(\frac{1}{8}b,\frac{1}{8}c\right)-\left(\frac{a}{b+c^{2}}+\max\left(\frac{1}{2}b,\frac{1}{2}c\right)\right)\ge 0$$ And $$\frac{a}{b+c^{2}}+\frac{b}{c+a^{2}}+\frac{c}{a+b^{2}}-\left(\frac{a}{b+c^{2}}+\frac{c}{a+b^{2}}+\min\left(\frac{1}{8}b,\frac{1}{8}c\right)\right)\ge0$$ And finally for $2\ge a\ge 1.6\geq c\ge 0.5\ge b$ : $$\frac{a}{b+c^2}+\max(0.5b,0.5c)\ge 1.5$$ and $a+b+c=3$ and $a,b,c\geq 0$
