Inequality

Let $a, \; b , \; c $ be real positive numbers that their product is equal to $1$. Prove that:

$$\frac{1}{a^3 (b+c)} +\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)} \geq \frac{3}{2}$$

Solution



We are using Tittu's lemma.

$$\begin{aligned}
\frac{1}{a^3 (b+c)} +\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)} &= \frac{\frac{1}{a^2}}{a(b+c)} + \frac{\frac{1}{b^2}}{b(a+c)}+ \frac{\frac{1}{c^2}}{c(b+a)} \\
 &\geq \frac{\left ( \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c} \right )^2}{2 (ab+bc+ca)} \\
 &= \frac{ab+bc+ca}{2}\\
 &\geq \frac{3}{2}
\end{aligned}$$