Minimum value of $\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$

Question

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.

Answer 1

I think it's correct.

By C-S we obtain: $$\sum_{cyc}\left(a+\frac{1}{b}\right)^2=\frac{1}{3}\sum_{cyc}1^2\sum_{cyc}\left(a+\frac{1}{b}\right)^2\geq$$ $$\geq\frac{1}{3}\left(\sum_{cyc}\left(a+\frac{1}{b}\right)\right)^2=\frac{1}{3}\left(12+\frac{1}{12}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right)^2\geq$$ $$=\frac{1}{3}\left(12+\frac{1}{12}\cdot9\right)^2=\frac{867}{16}.$$ The equality occurs for $a=b=c=4$, which says that $\frac{867}{16}$ is a minimal value.

Also we can use Jensen.

Let $f(x)=x^2+\frac{1}{x^2}$.

Hence, $f''(x)=2+\frac{6}{x^4}$, which says that $f$ is a convex function.

Thus, by Jensen and by AM-GM we obtain: $$\sum_{cyc}\left(a+\frac{1}{b}\right)^2=\sum_{cyc}\left(a^2+\frac{1}{a^2}\right)+2\sum_{cyc}\frac{a}{b}\geq$$ $$\geq3\left(\left(\frac{a+b+c}{3}\right)^2+\frac{1}{\left(\frac{a+b+c}{3}\right)^2}\right)+2\cdot3\sqrt[3]{\prod_{cyc}\frac{a}{b}}=$$ $$=3\left(16+\frac{1}{16}\right)+6=\frac{867}{16}.$$

Answer 2

Just another way: By QM-AM inequality, $$S=\sum \left(a+\frac1b\right)^2 \geqslant \frac13 \left(\sum a + \sum \frac1b\right)^2 \geqslant \frac13\left(12 + \frac34\right)^2$$