Školjka

Primijetimo da je $$\frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}=\frac {a^2b^2c^2}{a^{3}\left(b + c\right)} + \frac {a^2b^2c^2}{b^{3}\left(c + a\right)} + \frac {a^2b^2c^2}{c^{3}\left(a + b\right)}=\frac {b^2c^2}{a\left(b + c\right)} + \frac {c^2a^2}{b\left(c + a\right)} + \frac {a^2b^2}{c\left(a + b\right)}$$ Korištenjem $CSB$ nejednakosti dobivamo $$\left(a(b+c)+b(c+a)+c(a+b)\right)\left(\frac {b^2c^2}{a\left(b + c\right)} + \frac {c^2a^2}{b\left(c + a\right)} + \frac {a^2b^2}{c\left(a + b\right)}\right)\ge(ab+bc+ca)^2$$ $$\implies \frac {b^2c^2}{a\left(b + c\right)} + \frac {c^2a^2}{b\left(c + a\right)} + \frac {a^2b^2}{c\left(a + b\right)}\ge \frac{(ab+bc+ca)^2}{2(ab+bc+ca)}=\frac{ab+bc+ca}{2}\stackrel{AG}{\ge}\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}$$ Što je i trebalo dokazati.