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:
$$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2\ge 16$$
 A: 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 $1= ab+bc+ca\ge 3(abc)^{2/3}$  
A: 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?
A: By Cauchy-Schwarz inequality,
$$\begin{eqnarray*}
& & \left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2 \\
&=& \sqrt{\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2}\sqrt{\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(a+\frac{1}{b}\right)^2}\\
&\geq& \left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)+\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)+\left(c+\frac{1}{a}\right)\left(a+\frac{1}{b}\right)\\
&=&3+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+ab+bc+ca+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}.
\end{eqnarray*}$$
Now using the condition $ab+bc+ca = 1$, 
$$\begin{eqnarray*}
& & 3+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+ab+bc+ca+\frac{b}{a}+\frac{c}{b}+\frac{a}{c} \\
& = & \left(\frac{1}{ab}+9ab\right)+\left(\frac{1}{bc}+9bc\right)+\left(\frac{1}{ca}+9ca\right)+\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right) - 5 \\
& \geq & 6+6+6+3-5 = 16.
\end{eqnarray*}$$
We can also show that this inequality is strict, we can just put $(a, b, c) = \left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$.
(This minimizing case justifies our choice of $9ab+9bc+9ca$, combined with the equality condition of AM-GM inequality, and also of CS inequality.)
