Prove that ,$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2} \geq 8$

Question

For any positive a, b prove that $$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2} \geq 8$$

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My approach: Using the well known inequality,

$ \boxed{\mathrm{AM} \geq \mathrm{GM}}$

$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2} \geq 2 \sqrt{\left(a+\frac{1}{a}\right)^{2}*\left(b+\frac{1}{b}\right)^{2}}$

$\geq 2\left(a b+\frac{1}{a b}+\frac{a}{b}+\frac{b}{a}\right)$

what to do next?? Any hint or suggestion would be greatly appreciated

Answer 1

$x+\frac 1 x \geq 2$ for any positive number $x$ by AM-GM inequality. Hence $(a+\frac 1a)^{2}+(b+\frac 1 b)^{2} \geq 2^{2}+2^{2}=8$.

Answer 2

To continue your approach, just apply AM-GM a second time

$$\left(a+\frac{1}{a}\right)^{2}+\left(a+\frac{1}{a}\right)^{2} \geq 2 \sqrt{\left(a+\frac{1}{a}\right)^{2}\cdot\left(b+\frac{1}{b}\right)^{2}}$$ $$=2\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right) \stackrel{AM-GM}{\geq}2\cdot 2\sqrt{a\frac 1a}\cdot 2\sqrt{b\frac 1b}=8$$

Minimum is achieved for $a=b=1$.