Let the common difference be ( d > 0 ) (since terms are positive).

Then ( a_k = a_1 + (k-1)d )

Rationalize each term: [ \dfrac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \dfrac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k} = \dfrac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d} ]

The left-hand side becomes a telescoping sum: [ \sum_{k=1}^{n-1} \dfrac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \dfrac{1}{d} \sum_{k=1}^{n-1} \bigl( \sqrt{a_{k+1}} - \sqrt{a_k} \bigr) = \dfrac{1}{d} \bigl( \sqrt{a_n} - \sqrt{a_1} \bigr) ]

Now [ \dfrac{n-1}{\sqrt{a_1} + \sqrt{a_n}} = \dfrac{n-1}{\sqrt{a_1} + \sqrt{a_n}} \cdot \dfrac{\sqrt{a_n} - \sqrt{a_1}}{\sqrt{a_n} - \sqrt{a_1}} = \dfrac{(n-1)(\sqrt{a_n} - \sqrt{a_1})}{a_n - a_1} ]

But ( a_n - a_1 = (n-1)d ), so [ \dfrac{(n-1)(\sqrt{a_n} - \sqrt{a_1})}{(n-1)d} = \dfrac{\sqrt{a_n} - \sqrt{a_1}}{d} ]

Hence both sides are equal.

Proved.