$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}$ for $a,b,c\in\mathbb{R}^+$ with $abc=1$ Suppose that $a,b,c$ are positive reals such that $abc=1$. Prove that $$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}.$$
Hint: Use Titu's lemma.
My approach: I am trying to use Titu's lemma directly but it is not working. I meant that I wrote each term in the following way: $$\frac{a^3}{b+c}=\frac{a^3bc}{bc(b+c)}=\frac{a^2}{bc(b+c)}.$$ Then I applied Titu's lemma to the sum $$\frac{a^2}{bc(b+c)}+\frac{b^2}{ac(a+c)}+\frac{c^2}{ab(a+b)}\geq \frac{(a+b+c)^2}{bc(b+c)+ac(a+c)+ab(a+b)}=\frac{\sigma_1^2}{\sigma_1\sigma_2-3},$$ where $\sigma_1=a+b+c$ and $\sigma_2=ab+ac+bc$ are elementary symmetric functions.
This is what I got so far.
Would thankful if someone shows correct solution.

Notice that $$\sum_{cyc}\frac{a^3}{b+c}=\sum_{cyc}\frac{a^4}{a(b+c)}\geqslant \frac{(a^2+b^2+c^2)^2}{2(ab+bc+ca)}\geqslant \frac{a^2+b^2+c^2}{2}\geqslant \frac{3\sqrt[3]{abc}}{2}=\frac32 $$ Where we first used Titu's Lemma, then the well-known $a^2+b^2+c^2\geqslant ab+bc+ca$, and finally AM-GM.
