Say $ k$ of the $ a_i's$ are greater than or equal to $ 1$: $ a_1, a_2, \ldots, a_k \geq 1$. (Clearly, not all of them can be < 1). Then for $ i = 1, 2, \ldots, k$, the assertion of the problem is obvious. For $ i > k$, by AM-GM, $ \frac {a_1 + a_2 + \ldots + a_i}{i} \geq (a_1 \ldots a_i)^{1/i}$. But also, $ a_1 + a_2 + \ldots + a_i = \frac {1}{a_1} + \ldots + \frac {1}{a_n} - a_{i + 1} - a_{i + 2} - \ldots - a_{n} \geq \frac {1}{a_1} + \ldots + \frac {1}{a_i}$, because $ \frac {1}{x} - x = \frac {1 - x^2}{x} \geq 0 \text{ for } x < 1$. So by AM-GM, again, $ \frac {a_1 + a_2 + \ldots a_i}{i} \geq \frac {\frac {1}{a_1} + \ldots + \frac {1}{a_i}}{i} \geq \frac {1}{(a_1 a_2 \ldots a_i)^{1/i}}$. Since one of $ a_1 a_2 \ldots a_i$ or $ 1/(a_1 a_2 \ldots a_i)$ is bigger than or equal to $ 1$, we have the assertion.

The problem is very nice Due to the symmetry wlog assume that $ a_1\geq a_2\geq ...\geq a_k\geq 1\geq a_{k + 1}\geq...\geq a_n$ It is trivial that for $ j = 1,2,..,k$ we have that $ \sum_{i = 1}^j a_i\geq j$ Now observe that for $ j\geq k + 1$ we have that $ \sum_{i = 1}^j a_j\geq \sum_{i = 1}^j\frac {1}{a_j^2}$ (1) By Holder we have $ (\sum_{i = 1}^j a_j)^2(\sum_{i = 1}^j\frac {1}{a_j})\geq j^3$ so from (1) we conclude that $ \sum_{i = 1}^j a_j\geq j$ and we are done .

Sorry to everyone, I missed $ a_i^2$. Now problem is true...

Ahiles wrote: Sorry to everyone, I missed $ a_i^2$. Now problem is true... Fortunately the same proof I gave (and I think also the other proof using Cauchy-Schwartz) goes through.

jinjohn wrote: Fortunately the same proof I gave (and I think also the other proof using Cauchy-Schwartz) goes through. You should change it to Holder . See above my solution . I think it is true , isn't it ?

We may focus only on the numbers $ a_i$ which are smaller than $ 1$, but here is another aproach: It suffices to prove that $ a_1+a_2+\cdots a_n\geq n$. We shall first prove this assertion. We have \[ \sum_{i=1}^{n} a_i +n=\sum_{i=1}^{n}(\frac{1}{a_{i}^{2}}+1)\geq 2\sum_{i=1}^{n} \frac{1}{a_{i}}.\]By adding $ 2\sum_{i=1}^{n} a_i$ in both terms, we obtain \[{ 3\sum_{i=1}^{n} a_i +n\geq 2\sum_{i=1}^{n}(a_i+\frac{1}{a_i}})\geq 4n\] Therefore $ \sum_{i=1}^{n} a_i \geq n$. Now assume that for some $ k$, we have that \[ \displaystyle \sum_{m=1}^{k} a_{i_m}< k,\] for any $ \{i_1, i_2\dots i_k\}\subset\{1, 2\dots n\}$. Now we add all these possible sums together and we obtain \[ \ds \binom{n-1}{k-1}\cdot \sum_{i=1}^{n} a_i< k\cdot\binom{n}{k}.\]By simplifying we obtain $ \sum_{i=1}^{n} a_i<n$, which is a contradiction. PS. This problem was proposed by me and Georgescu Flavian

（a1+...+an)^3=（a1+...+an)^2 * (1/a1^2+...+1/an^2)\gep n^3 we have a1+...+an\gep n