Problem. Let $a,b,c>0: \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3.$ Prove that: $$\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\ge\frac{3}{2}\left(\frac{a+b+c}{ab+bc+ca}\right)^2$$
I solved the problem, but my proof is much calculating work. I hope to see other ideas.
We can rewrite the inequality as homogenous form: $$\sum_{cyc}\frac{a}{b^{2}+c^{2}}\geq \frac{9abc(a+b+c)^{2}}{2(bc+ca+ab)^{3}}.$$ By C-S $$LHS\geq \frac{\displaystyle \left(\sum_{cyc}a^{2}\right)^{2}}{\displaystyle \sum_{cyc}a^{3}\left(b^{2}+c^{2}\right)}.$$ It remains to show that: $$2\left(\sum_{cyc}a^{2}\right)^{2}\left(\sum_{cyc}bc\right)^{3}\geq 9abc(a+b+c)^{2}\left[\sum_{cyc}b^{2}c^{2}(b+c)\right].\quad (*)$$ Rewrite it in $pqr$ form: $$f(r)=9p^{2}(2p^{2}+q)r^{2}-9p^{3}q^{2}r+2q^{3}(p^{2}-2q)^{2}\geq 0.$$ We need to check two cases:
Case 1: $r=\frac{pq^{2}}{4p^{2}+2q}$.
By Schur's inequality degree $3$, we have $\frac{pq^{2}}{4p^{2}+2q}\geq \frac{p(4q-p^{2})}{9}$,
or $p^{2}\geq \frac{7+3\sqrt{5}}{4}q$ This yields $\Delta_{r}<0$, we have $Q.E.D$.
Case 2: $r\neq \frac{pq^{2}}{4p^{2}+2q}$.
In this case, $f(r)$ reaches its minimum at its extreme point, which means two of $a,b,c$ are equal.
WLOG, assume that $b=c=t$. Then we have $$9p^{2}(2p^{2}+q)r^{2}-9p^{3}q^{2}r+2q^{3}(p^{2}-2q)^{2}=2t^{3}(a+t)\left(8a^{4}+11ta^{3}+12t^{2}a^{2}-8t^{3}a+4t^{4}\right)(a-t)^{2}\geq 0.$$ The proof is completed.
