let $a,b,c>0$ and such $ab+bc+ca=3$,show that $$\dfrac{a}{a^3+b^2+1}+\dfrac{b}{b^3+c^2+1}+\dfrac{c}{c^3+a^2+1}\le 1$$
I want use Cauchy-Schwarz inequality $$(a^3+b^2+1)(\frac{1}{a}+1+c^2)\ge (a+b+c)^2$$ so $$\sum_{cyc}\dfrac{a}{a^3+b^2+1}\le\sum_{cyc}\dfrac{1+a+ac^2}{(a+b+c)^2}$$ it must to prove $$(a+b+c)^2\ge \sum_{cyc}(1+a+ac^2)=3+(a+b+c)+ac^2+ba^2+cb^2$$ then I can't do it.
The inequality, which you got is wrong.
Try $c\rightarrow0^+$ and $a=2$.
My proof.
Since $$a^3+b^2+1\geq a^2+b^2+a,$$ it's enough to prove that $$\sum_{cyc}\frac{a}{a^2+b^2+a}\leq1$$ or $$\sum_{cyc}\left(a^4b^2+a^4c^2-a^3b-a^2bc+\frac{2}{3}a^2b^2c^2-\frac{2}{3}abc\right)\geq0$$ or
$$\sum_{cyc}\left(3a^4b^2+3a^4c^2-a^4b^2-a^4bc-a^3b^2c-a^3b^2c-a^3c^2b-a^2b^2c^2+2a^2b^2c^2-2abc\right)\geq0$$ or $$\sum_{cyc}\left(2a^4b^2+3a^4c^2-a^4bc-2a^3b^2c-a^3b^2c+a^2b^2c^2-2abc\right)\geq0$$ and since $$\sum_{cyc}a^4c^2\geq\sum_{cyc}a^3b^2c,$$ $$\sum_{cyc}(2a^4b^2+2a^4c^2)\geq\sum_{cyc}4a^4bc$$ and by Schur $$\sum_{cyc}(a^4bc-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0,$$ it's enough to prove that $$\sum_{cyc}(a^4bc-abc)\geq0$$ or $$a^3+b^3+c^3\geq3,$$ which is true by Power Mean: $$a^3+b^3+c^3\geq\sqrt{\frac{(a^2+b^2+c^2)^3}{3}}\geq\sqrt{\frac{(ab+ac+bc)^3}{3}}=3.$$