obviously: 2*(a1)^2 + 2*(a1)*S + (S^2 + T) = 101, where S = a2+a3...+a100 T = a2^2 + a3^2 ... + a100^2. since we also know T >= S^2 / 99, S^2 * (100/99) + S*(2*a1)+(2*a1^2 - 101) >= 0. so delta = 4*a1^2 - 4* (100/99) * (2*a1^2 - 101) >= 0. so a1^2 > 200/99*a1^2 - 101*100/99 and 101*100/99 > 101/99*a1^2 and |a1| <= 10

Hmm.. vinoth didn't you post the solution 11 or 12 times already? what happened with this site? Palosh alias Downito _______________________________ "Our days are never coming back... " (Highway Song -> System of a down)

I see this thread has 11 more posts than it should

crap im very sorry im such an idiot my internet connection was very slow and the page didn't load when i pressed submit ... so i think i must have pressed it again not knowing it would actually submit twice , etc.

the server was down for a couple of hours this morning, that is why the site was going slow.

Here is a nice solution for this nice problem. Let us assume by contradiction that $|a_{k}| >10,$ for some $k$. WLOG, let $k=1.$ So $a_{1}^{2} >100.$ and $a_{2}^{2}+a_{3}^{2}+\ldots+a_{100}^{2}+s^{2} <1,$ where $s=a_{1}+a_{2}+...+a_{n}.$ But, from Cauchy-Schwarz inequality we have: \[ a_{1}^{2}\leq 100(a_{2}^{2}+a_{3}^{2}+\ldots+a_{100}^{2}+s^{2}) <100. \] We get a contradiction.

Solution similar to arqady's one posted here: http://www.mathlinks.ro/Forum/viewtopic.php?t=66908 WLOG, lets assume that $max\{|a_1|,...,|a_{100}|\}=|a_{100}|.$ So, we have that: \begin{eqnarray*}101 &=& a_1^2+...+a_{99}^2+a_{100}^2 + (a_1+ ...+ a_{99})^2 + a_{100}^2 + 2a_{100}(a_1+ ...+ a_{99})\\ &=&2a_{100}^2 + 2a_{100}(a_1+...+a_{99})+(a_1+...+a_{99})^2+a_1^2+...+a_{99}^2\end{eqnarray*} By Cauchy, we get that: $(a_1^2+...+a_{99}^2)(1^2+...+1^2)\geq (a_1+...+a_{99})^2\Rightarrow a_1^2+...+a_{99}^2\geq \frac{(a_1+...+a_{99})^2}{99}$ So: $2a_{100}^2 + 2a_{100}(a_1+...+a_{99})+\frac{100}{99}(a_1+...+a_{99})^2-101\leq 0$ Let $a_1+...+a_{99} = s$, so we have that: $\frac{100}{99}t^2+2a_{100}t+2a_{100}^2-101\leq 0$ Finally, $\Delta = 4a_{100}^2-\frac{400}{99}(2a_{100}^2-101)\geq 0$ from what we get that $\boxed{|a_{100}|\leq \sqrt{100}=10}$ $\blacksquare$

Cezar Lupu wrote: But, from Cauchy-Schwarz inequality we have: \[ a_{1}^{2}\leq 100(a_{2}^{2}+a_{3}^{2}+\ldots+a_{100}^{2}+s^{2}) <100. \]We get a contradiction. I think your reasoning was (1+1+…+1)(a_2^2+a_3^2+…+a_100^2+s^2) >= (a_2+a_3+…+a_100+s)^2 >= (a_1)^2, where the last step follows from |2a_2+2a_3+…+2a_100+a_1| >= |a_1|, but this could not be true if a_2+a_3+…+a_100 has opposite sign of a_1, as it would make the first expression closer to 0 than a_1 is. @above nice solution! A stronger version means (a_100)^2 <= 101/2, or a_100 <= +- sqrt(101/2).