l don't know what telescopic sum is...it's a classical sum ! i'll give you the solution so maybe it will help. let $x_k=\sqrt{a_k}-\sqrt{a_{k=1}}$ for $k=1,\ldots,n$ so $a_1=(x_1+\cdots+x_n)^2, a_2=(x_2+\cdots+x_n)^2,\ldots,a_n=x_n^2$ after having developped the squares, we get \[ \sum_ {k=1}^n a_k=\sum_{k=1}^n k x_k^2+2 \sum_{1\leq k<\leq n} k x_k x_l (1)\] $x_k x_l$'s coefficient ($k<l$) in the last sum is equal to $k$ because this product appears exactly one time in each expression of $a_1,\ldots,a_n$ but isn't present in $a_{k+1},\ldots,a_n$ the square on the right member of the inequality equals \[ \left( \sum_{k=1}^n \sqrt k x_k \right)^2= \sum_{k=1}^n k x_k^2+2\sum_{1 \leq k<l \leq n} \sqrt{kl} x_k x_l. (2)\] since the value of (1) isn't $\geq$ than the value of (2), the result follows.

- $\displaystyle \sum_{k=1}^n (\sqrt{a_k}-\sqrt{a_{k+1}})=(\sqrt{a_1}-\sqrt{a_2})+(\sqrt{a_2}-\sqrt{a_3})+\cdots+(\sqrt{a_n}-\sqrt{a_{n+1}})$ - $a_1\geq a_2 \geq \ldots \geq a_2 \geq a_{n+1}=0$ that's all what l can tell you. howewer, the fact that $a_{n+1}=0$ doesn't intervene in the solution written above.

now, if you really meant it that way, then what i've written first is a solution! we have $\sum^n_{k=1} (\sqrt{a_k}-\sqrt{a_{k+1}})=\sqrt{a_1}-\sqrt{a_2}+\sqrt{a_2}\pm...-\sqrt{a_{n+1}}$, so all terms except $+\sqrt{a_1}-\sqrt{a_{n+1}}$ cancel, so the right side of your inequality is just $\sqrt{a_1}$. on the left side we get the square root of $a_1$ plus some other nonnegative terms, so the inequality is true. Peter [Edit: wrong index]

l gave the official solution but yours is shorter and smarter, peter : congratulations. therefore, this solution is very easy, and the problem is an imo problem ! (they have become harder in the future)

Lemma: If $k \geq l$, then $\sqrt{k+d} - \sqrt{k} \leq \sqrt{l+d} - \sqrt{l}$. Proof of lemma: Just expand it lol Now, rearranging the inequality gives $\sqrt{a_1+a_2+\ldots+a_n} - \sqrt{a_1} \leq \sqrt{2a_2}-\sqrt{a_2}+\sqrt{3a_3}-\sqrt{2a_3}+\ldots+\sqrt{ka_k}-\sqrt{(k-1)a_k}$. Now, repeatedly applying the lemma gives $\sqrt{2a_2}-\sqrt{a_2} \geq \sqrt{a_1+a_2}-\sqrt{a_1}$, $\sqrt{3a_3}-\sqrt{2a_3} \geq \sqrt{a_1+a_2+a_3}-\sqrt{a_1+a_2}$, and so on. The sum telescopes, giving the desired inequality.

Is this in an older IMO?