<s>[INST] Minimum value of $\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$ 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.
 [/INST] Just another way: By QM-AM inequality,
$$S=\sum \left(a+\frac1b\right)^2 \geqslant \frac13 \left(\sum a + \sum \frac1b\right)^2 \geqslant \frac13\left(12 + \frac34\right)^2$$
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