Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 60

Show that $\displaystyle 2 \sqrt{x} > 3 - \frac{1}{x}$ for all $x > 1$.

If $\displaystyle 2\sqrt{x} > 3 - \frac{1}{x}$ for all $x > 1$, then $f(x)$ must be increasing on $(1 , \infty)$
So,
$\displaystyle f(x) = 2\sqrt{x} - 3 + \frac{1}{x}$
By taking the derivative

$
\begin{equation}
\begin{aligned}
f'(x) &= 2 \left( \frac{1}{2\sqrt{x}} \right) - \frac{1}{x^2}\\
\\
f'(x) &= \frac{1}{\sqrt{x}} - \frac{1}{x^2}
\end{aligned}
\end{equation}
$


We can say that $f(x)$ is increasing when, $f'(x) > 0$, $\displaystyle \frac{1}{\sqrt{x}} > \frac{1}{x^2}$
We have $\displaystyle \frac{x^2}{\sqrt{x}} > 1$ which implies for all $x > 1$
Therefore, $\displaystyle 2 \sqrt{x} > 3 - \frac{1}{3}$ for all $x > 1$