Given $a,b,c \in \mathbb{R^+}$ such that $a+b+c=12$
Find Minimum value of $$S=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$$
My Try: By Cauchy Schwarz Inequality we have
$$\left(a+\frac{1}{b}\right)+\left(b+\frac{1}{c}\right)+\left(c+\frac{1}{a}\right)\le \sqrt{3}\sqrt{S}$$
$\implies$
$$\sqrt{3S} \ge 12+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
Now by $AM \ge HM $ inequality we have
$$\frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ $\implies$
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{3}{4}$$ hence
$$\sqrt{3S} \ge 12+\frac{3}{4}=\frac{51}{4}$$
hence
$$3S \ge \frac{2601}{16}$$
so $$S \ge \frac{867}{16}$$
is this approach correct and any better approach please share.