Question 64: Given $a, b \in [1, \sqrt{3}]$, find the maximum and minimum values of $\frac{a^{2}+b^{2}-1}{a b}$.

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Question 64, Solution: First, we know that $\frac{a^{2}+b^{2}-1}{a b} \geq \frac{2 a b-1}{a b}=2-\frac{1}{a b} \geq 2-1=1$, with equality holding when $a=b=1$.

On the other hand, assuming without loss of generality that $a \leq b$, then $\frac{a^{2}+b^{2}-1}{a b}=\frac{1}{a}\left(b+\frac{a^{2}-1}{b}\right) \leq \frac{1}{a}\left(\sqrt{3}+\frac{a^{2}-1}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}\left(a+\frac{2}{a}\right) \leq$ $\frac{1}{\sqrt{3}}\left(1+\frac{2}{1}\right)=\sqrt{3}$, with equality holding when $a=1, b=\sqrt{3}$.