Prove that $\frac{a^3}{(1+b)(1+c)}+\frac{b^3}{(1+a)(1+c)}+\frac{c^3}{(1+a)(1+b)} \geq \frac{3}{4}.$

Question

Let $a,b,$ and $c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{a^3}{(1+b)(1+c)}+\dfrac{b^3}{(1+a)(1+c)}+\dfrac{c^3}{(1+a)(1+b)} \geq \dfrac{3}{4}.$$

Attempt

We have $$\dfrac{a^3}{(1+b)(1+c)}+\dfrac{b^3}{(1+a)(1+c)}+\dfrac{c^3}{(1+a)(1+b)} = \dfrac{a^3}{1+b+c+bc}+\dfrac{b^3}{1+a+b+ac}+\dfrac{c^3}{1+a+b+ab} = \dfrac{a^4}{a+ab+ac+1}+\dfrac{b^3}{b+ab+bc+1}+\dfrac{c^4}{c+bc+ac+1}.$$

I get stuck here.

Answer 1

From the AM-GM inequality :

$$\frac{a^3}{(1+b)(1+c)}+\frac{1+b}{8}+\frac{1+c}{8} \geq 3\sqrt[3]{\frac{a^3}{(1+b)(1+c)} \cdot \frac{1+b}{8}\cdot\frac{1+c}{8}}=\frac{3a}{4}$$

Now do this for the others and add them all to get :

$$\frac{a^3}{(1+b)(1+c)}+\frac{b^3}{(1+a)(1+c)}+\frac{c^3}{(1+a)(1+b)}\geq \frac{a+b+c}{2} -\frac{3}{4}\geq \frac{3}{4}$$ because : $$a+b+c \geq 3\sqrt[3]{abc}=3$$ from AM-GM.