1995 IMO #2

 Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that \[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}. \]

Solution 1: 

We have $$\sum_{\text{cyc}}\frac{1}{a^3(b+c)}=\sum_{\text{cyc}}\frac{(bc)^3}{b+c}=\sum_{\text{cyc}}\frac{(bc)^2}{\frac{b+c}{bc}}=\sum_{\text{cyc}}\frac{(bc)^2}{\frac{1}{b}+\frac{1}{c}}.$$By Titu's Lemma, we obtain that$$\sum_{\text{cyc}}\frac{(bc^2)}{\frac{1}{b}+\frac{1}{c}} \geq \frac{(ab+bc+ca)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}=\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)} \qquad (\clubsuit).$$However, note that by AM-GM,$$ab+bc+ac \geq 3\sqrt[3]{(abc)^2}=3 \implies \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 3$$$$\implies \frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)} \geq 3$$$$\implies \frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)} \geq \frac{3}{2},$$which, combined with $(\clubsuit),$ gives us the result. $\square$

Solution 2: 

By Cauchy-Schwarz, we know that$$\left(\sqrt{\frac{1}{a^3(b+c)}}^2 + \sqrt{\frac{1}{b^3(a+c)}}^2 + \sqrt{\frac{1}{b^3(a+b)}}^2 \right) \left(\sqrt{a(b+c)}^2 + \sqrt{b(a+c)}^2 + \sqrt{c(a+b)}^2\right) \ge \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^2.$$Using the fact that $abc = 1$, we know that $\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)^2 = (ab + bc + ca)^2$. So from the Cauchy-Schwarz inequality, we know that$$\left(\sqrt{\frac{1}{a^3(b+c)}}^2 + \sqrt{\frac{1}{b^3(a+c)}}^2 + \sqrt{\frac{1}{b^3(a+b)}}^2 \right) \left(\sqrt{a(b+c)}^2 + \sqrt{b(a+c)}^2 + \sqrt{c(a+b)}^2\right) \ge \frac{(ab+ bc + ac)^2}{2}.$$By AM-GM and the condition that $abc = 1$, we know that $\frac{(ab+ bc + ac)^2}{2} \ge \frac{3}{2}.$ Thus$$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \ge \frac{3}{2}.$$