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Find the minimum value of \(\dfrac{a}{b} + \dfrac{b}{a}\), where a and b are positive real numbers.

\(\mathbf{\huge{AM \geq GM }}\)

\(\begin{array}{|rcll|} \hline \dfrac{a^2+b^2}{2} &\geq& \sqrt{a^2b^2} = ab \\ \dfrac{a^2+b^2}{2} &\geq& ab \quad | \quad * \dfrac{2}{ab} \\ \dfrac{a^2+b^2}{ab} &\geq& 2 \\ \dfrac{a^2}{ab}+\dfrac{b^2}{ab} &\geq& 2 \\ \mathbf{\dfrac{a}{b}+\dfrac{b}{a}} &\geq& \mathbf{2} \\ \hline \end{array}\)

The minimum value of \(\dfrac{a}{b} + \dfrac{b}{a}\) is 2