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:
$$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2\ge 16$$

This is problem J57 of mathematical reflections, vol.2007, issue 4; 
$a^2+b^2+c^2 \ge ab+bc+ca =1$
$2\left(\frac ab+\frac bc+\frac ca\right)\ge6$ (as BenjaLim does)
It remains 
$\frac 1{a^2} + \frac 1{b^2} + \frac 1{c^2} \ge 9$ which follows by (use AGM) $\frac 3{(abc)^{2/3}}\ge9$ or $(abc)^{2/3}\le \frac13.$ This is a consequence of $ab+bc+ca=1$ since always by AGM $1= ab+bc+ca\ge 3(abc)^{2/3}$  
