Inequality $(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2\ge 16$ For every real positive number $a,b,c$ such that $ab+bc+ca=1$, how to prove that:
$$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2\ge 16$$

Expanding the left hand side, we find that because for all non-zero real numbers $x$, $x^2 + x^{-2} \geq 2$ that
$$\left(a + \frac{1}{a}\right)^2 + \ldots \left(c + \frac{1}{c}\right)^2 \geq \frac{2a}{b} + \frac{2b}{c} + \frac{2c}{a} + 6.$$
Can you complete the argument?
