\(\begin{aligned}
f(x)&=\dfrac{x^2-4+36}{4(x-2)}\\
&=\dfrac{x^2-4}{4(x-2)}+\dfrac{36}{4(x-2)}\\
&=\dfrac{x+2}{4}+\dfrac{9}{x-2}\\
&=\dfrac{(x-2)+4}{4}+\dfrac{9}{x-2}\\
&=\dfrac{x-2}{4}+1+\dfrac{9}{x-2}.
\end{aligned}\)

Áp dụng bất đẳng thức Cauchy cho hai số dương \(\dfrac{x-2}{4}\) và \(\dfrac{9}{x-2}\) ta có $$\begin{eqnarray*}
&\dfrac{x-2}{4}+\dfrac{9}{x-2}&\geq2\sqrt{\dfrac{x-2}{4}\cdot\dfrac{9}{x-2}}\\
\Leftrightarrow&\dfrac{x-2}{4}+\dfrac{9}{x-2}&\geq3\\
\Leftrightarrow&\dfrac{x-2}{4}+\dfrac{9}{x-2}+1&\geq4\\
\Leftrightarrow&f(x)&\geq4.
\end{eqnarray*}$$