Question

Find the min value:

${(x+\frac{1}{x})}^2$

Answer 1

Let us consider the terms $x$ and $\frac{1}{x}$

$AM\ge GM\frac{x+\frac{1}{x}}{2}\ge \sqrt{x\frac{1}{x}}x+\frac{1}{x}\ge 2$

So the min value of $x+\frac{1}{x}$ is $2$

The min value of ${(1+\frac{1}{x})}^2$ is $2^2=4$