Induction works!

Full proof?

See here http://www.artofproblemsolving.com/Forum/viewtopic.php?p=23120&sid=0ecfd9a96555096cbce4cb53167d61d4#p23120

For $ n=1 $, the inequality is trivial. Suppose the statement is true for $ n=1,2...,m $, we consider $ n=m+1 $ now. First, denote $ k=\sqrt{a_1^2+a_2^2+...+a_m^2} $ then : $ LHS = \sqrt{((k-\sqrt{n}a_{n+1})+\sqrt{n}a_{n+1})^2+a_{n+1}^2} \le k-\sqrt{n}a_{n+1}+\sqrt{n+1}a_{n+1} $ $ \le (a_1+\frac{a_2}{\sqrt{2}+1}+\cdots+\frac{a_n}{\sqrt{n}+\sqrt{n-1}})+\frac{a_{n+1}}{\sqrt{n+1}-\sqrt{n}} = RHS $, done!!

Let $a_{n+1}=0$ and $b_{k}=\sqrt{k}(a_{k}-a_{k+1}),$ then $b_{k}\ge 0$ and $a_{j}=\frac{b_{k}}{\sqrt{k}}+\frac{b_{k+1}}{\sqrt{k+1}}+...+\frac{b_{n}}{\sqrt{n}}.$ Now we just expand in terms of the $b_{k}$'s, we get \[LHS^{2}=\sum_{k=1}^{n}(\sum_{j=k}^{n}\frac{b_{j}}{\sqrt{j}})^{2}=b_{1}^{2}+b_{2}^{2}+...+b_{n}^{2}+2\sum_{k<l}\frac{\sqrt{k}}{\sqrt{l}}b_{k}b_{l}\]\[\leq \sum_{k}b_{k}^{2}+2\sum_{k<l}b_{k}b_{l}=(b_{1}+b_{2}+...+b_{n})^{2}=RHS^{2}\]as desired. Now, equality holds when then at least $n-1$ of the numbers $b_{1},b_{2},...,b_{n}$ are zeros, but since $b_{n}>0,$ we see that $b_{1}=b_{2}=...=b_{n-1}=0,$ and so $a_{1}=a_{2}=...=a_{n}$ is the only equality case.