Solution:

By Cauchy-Schwarz, $\left(\sum x_{i}^{1/2}\right)^{2} \leq \left(\sum 1\right)\left(\sum x_{i}\right)$, with equality if and only if all $x_{i}$ are equal. In other words, if we put $x_{n+1} = \dfrac{x_{1} + x_{2} + \ldots + x_{n}}{n}$, then $\sum x_{i}^{1/2} \leq n x_{n+1}^{1/2}$.

But since all $x_{i} \geq 1$, we have

$x_{1}^{1/2} + x_{2}^{1/3} + x_{3}^{1/4} + \ldots + x_{n}^{1/(n+1)} \leq \sum x_{i}^{1/2}$

with equality if and only if $x_{2} = x_{3} = \ldots = x_{n} = 1$.

Hence

$x_{1}^{1/2} + x_{2}^{1/3} + x_{3}^{1/4} + \ldots + x_{n}^{1/(n+1)} \leq x_{n+1}^{1/2}$

with equality if and only if all $x_{i} = 1$.