Using the Arithmetic Mean-Geometric Mean Inequality Let a, b, c be positive real numbers. Prove that
$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq \sqrt{3(a^{2}+b^{2}+c^{2})}$

Rewrite as $a^2b^2 + b^2c^2 + c^2a^2 \ge \sqrt{3(a^4b^2c^2 + a^2b^4c^2 + a^2b^2c^4)}$
Let $x = a^2b^2, y = b^2c^2, z = c^2a^2$.
Therefore, now we have to prove : $x + y + z \ge \sqrt{3(xy + yz + zx)}$
Squaring both sides, $(x + y + z)^2 \ge 3(xy + yz + zx)$
$x^2 + y^2 + z^2 \ge xy + yz + zx$.
Now, use A.M. $\ge$ G.M. $(\frac{x^2 + y^2}2 \ge xy)$
