$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}$ 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.
 A: 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.
A: Also, by Holder and AM-GM $$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(b+c)}=\frac{1}{6}(a+b+c)^2\geq\frac{9}{6}=\frac{3}{2}.$$
A: Also, we can use AM-GM and SOS:
$$\sum_{cyc}\frac{a^3}{b+c}=\frac{a^2+b^2+c^2}{3}+\sum_{cyc}\left(\frac{a^3}{b+c}-\frac{a^2}{2}\right)\geq$$
$$\geq\frac{3}{2}+\frac{1}{2}\sum_{cyc}\frac{a^2(2a-b-c)}{b+c}=\frac{3}{2}+\frac{1}{2}\sum_{cyc}\frac{a^2(a-b-(c-a))}{b+c}=$$
$$=\frac{3}{2}+\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{a^2}{b+c}-\frac{b^2}{c+a}\right)=\frac{3}{2}+\sum_{cyc}\tfrac{(a-b)^2(a^2+b^2+ab+ac+bc)}{2(a+c)(b+c)}\geq\frac{3}{2}.$$
The following stronger inequality is also true.

Let $a$, $b$ and $c$ be positive numbers such that $a^5+b^5+c^5=3.$ Prove that:
$$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq\frac{3}{2}.$$

