Question

Solve the following inequality:
$\sqrt{x+\frac{1}{x^2}}+\sqrt{x-\frac{1}{x^2}}>\frac{2}{x}.$

Answer 1

$\sqrt{x+\frac{1}{x^2}}+\sqrt{x-\frac{1}{x^2}}>\frac{2}{x}$

now$\frac{x-1}{x^2}>0x^3>0$

squaring both sides

$x+\frac{1}{x^2}+x-\frac{1}{x^2}+2\sqrt{x^2-\frac{1}{x^4}}>\frac{4}{x^2}2x+2\sqrt{x^2-\frac{1}{x^4}}>\frac{4}{x^2}\sqrt{x^2-\frac{1}{x^4}}>\frac{2}{x^2}-x$

squaring again

$x^2-\frac{1}{x^4}>\frac{4}{x^4}+x^2-\frac{2}{x}\Rightarrow \frac{2}{x}>\frac{5}{x^4}\Rightarrow x^3>\frac{5}{2}x>\sqrt[3]{3\frac{5}{2}}$.