Prove that $\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}$

My question: Given $$a,b,c$$ are 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!

• Comments are not for extended discussion; this conversation has been moved to chat. May 27, 2021 at 19:56

By C-S and by the Vasc's inequality we obtain:$$\sum_{cyc}\frac{a}{b+c^2}=\sum_{cyc}\frac{a^3}{a^2b+a^2c^2}\geq\frac{\left(\sum\limits_{cyc}\sqrt{a^3}\right)^2}{\sum\limits_{cyc}(a^2b+a^2b^2)}=$$ $$=\frac{3\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}a^2b\sum\limits_{cyc}a+3\sum\limits_{cyc}a^2b^2}=\frac{3\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}(4a^2b^2+a^3b+a^2bc)}\geq$$ $$\geq\frac{3\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}(4a^2b^2+a^2bc)+\frac{1}{3}(a^2+b^2+c^2)^2}=\frac{(a+b+c)\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}(4a^2b^2+a^2bc)+\frac{1}{3}(a^2+b^2+c^2)^2}$$ and it's enough to prove that: $$\frac{(a+b+c)\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}(4a^2b^2+a^2bc)+\frac{1}{3}(a^2+b^2+c^2)^2}\geq\frac{3}{2}$$ or $$\sum_{cyc}\left(a^{4}+2a^{3}b+2a^{3}c+4\sqrt{a^5b^3}+4\sqrt{a^5c^3}+4\sqrt{a^3b^3c^2}-14a^{2}b^{2}-3a^{2}bc\right)\geq0,$$ which is smooth.