$a,b,c > 0$ and $a+b+c=1$. Prove that
$$\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\geqslant \frac{13}{2(1-abc)} $$
This inequality can be solved by computer ( essentially, remove the constrain through homogenizing, clear denominator, expanding and proving positive terms). Since this mechanical strategy is very simple to do but impractical during math contest, I am looking forward to solution using classical inequalities approach.
I would not call this an answer, but perhaps could be helpful. Observe that: \begin{align*} (1-a)(1-b)(1-c)&=\color{red}{1-a-b-c}+ab+ac+bc-abc\\ &=ab+ac+bc-abc \\ &=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}-abc \\ &=\frac{1-(a^2+b^2+c^2)}{2}-abc \\ &=1-abc-\frac{1+a^2+b^2+c^2}{2} \\ &=1-abc-\frac{a+b+c+a^2+b^2+c^2}{2}. \end{align*} Thus, $$a+b+c+a^2+b^2+c^2=2[(1-abc)-(1-a)(1-b)(1-c)].$$ By Cauchy Schwartz inequality, $$(a+b+c+a^2+b^2+c^2)\left(\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\right)\geq3^2=9.$$ As a result, $$\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\geq\frac{9}{2[(1-abc)-(1-a)(1-b)(1-c)]}.$$ If we can show $$\frac{9}{2[(1-abc)-(1-a)(1-b)(1-c)]}\geq\frac{13}{2(1-abc)},$$ then we are done. Suppose this is true. It gives $$13[(1-abc)-(1-a)(1-b)(1-c)]\leq 9(1-abc),$$ or equivalently, $$(1-abc)\leq\frac{13}{4}(1-a)(1-b)(1-c).$$ If this inequality is true under the constraint $a+b+c=1$ and $a,b,c>0$, then we are done. I checked that the equality precisely holds when $a=b=c=1/3$. Hope anyone can finish it.
