We prove the inequality by induction on $n \geq 2$.

For $n=2$, let $a_{1} \geq a_{2}>0$. We have
$$\begin{aligned} \sqrt{a_{1}^{2}+a_{2}^{2}} & =\sqrt{2 a_{2}^{2}+\left(a_{1}-a_{2}\right)^{2}+2 a_{2}\left(a_{1}-a_{2}\right)} \\ & \leq \sqrt{2 a_{2}^{2}+\left(a_{1}-a_{2}\right)^{2}+2 \sqrt{2} a_{2}\left(a_{1}-a_{2}\right)}=\left(a_{1}-a_{2}\right)+\sqrt{2} a_{2} \\ & \leq a_{1}+\frac{a_{2}}{\sqrt{2}+1} \end{aligned}$$

Assume the inequality true for any non increasing sequence of positive $n$ real numbers and let $a_{1} \geq a_{2} \geq \cdots \geq a_{n+1}>0$. Denote $q_{n}=\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}$. It is clear that $q_{n} \geq \sqrt{n} a_{n+1}$. We have
$$\begin{aligned} \sqrt{a_{1}^{2}+a_{2}^{2}} & +\cdots+a_{n+1}^{2}=\sqrt{q_{n}^{2}+a_{n+1}^{2}}= \\ & =\sqrt{(n+1) a_{n+1}^{2}+\left(q_{n}-\sqrt{n} a_{n+1}\right)^{2}+2 \sqrt{n} a_{n+1}\left(q_{n}-\sqrt{n} a_{n+1}\right)} \\ & \leq \sqrt{(n+1) a_{n+1}^{2}+\left(q_{n}-\sqrt{n} a_{n+1}\right)^{2}+2 \sqrt{n+1} a_{n+1}\left(q_{n}-\sqrt{n} a_{n+1}\right)} \\ & \leq \sqrt{n+1} a_{n+1}+\left(q_{n}-\sqrt{n} a_{n+1}\right)=q_{n}+\frac{a_{n+1}}{\sqrt{n+1}+\sqrt{n}} \\ & \leq a_{1}+\frac{a_{2}}{\sqrt{2}+1}+\cdots+\frac{a_{n+1}}{\sqrt{n+1}+\sqrt{n}} \end{aligned}$$

The equality holds when $q_{i}=\sqrt{i} a_{k+1}$, for $i=1,2, \ldots, n-1$, that is $a_{1}= a_{2}=\cdots=a_{n}$.