If $a_1=-1$ and $x_2=x_3=...=x_n\rightarrow0^+$ we get something wrong. Hence, all $a_i=1$ and we still get something wrong. Or you mean that $x_i$ are given?

arqady wrote: Or you mean that $x_i$ are given? Yes for each set of $x_i$, we have a set of $a_i$. In other words $a_i$'s are dependent of $x_i$'s.

arqady wrote: If $a_1=-1$ and $x_2=x_3=...=x_n\rightarrow0^+$ we get something wrong. Hence, all $a_i=1$ and we still get something wrong. Or you mean that $x_i$ are given? Dear arqady Sayan's formulation is very clear, in my opinion.

If $x_1\geq x_2\geq...\geq x_n>0$ then $\sum_{k=1}^n(-1)^{k+1}x_k^2\geq\left(\sum_{k=1}^n(-1)^{k+1}x_k\right)^2$. Nice problem! If $n$ is an odd number then we can use induction. A case with $n$ is an even number follows from the previous case.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2456684&sid=b8c3332752208f22f2eb32bc12f33fea#p2456684

This seems too simple, did I mess something up?

yofro wrote: This seems too simple, did I mess something up? $$\left(\sum a_ix_i^2\right)\left(\sum a_i\right)\neq \left(\sum a_ix_i^2\right)$$

WolfusA wrote: yofro wrote: This seems too simple, did I mess something up? $$\left(\sum a_ix_i^2\right)\left(\sum a_i\right)\neq \left(\sum a_ix_i^2\right)$$ That's not what I was suggesting... $\sqrt{a_i}\sqrt{a_i}x_i=a_i*x_i$.

Thanks to @sky.mty, the god of algebra! First, we sort the sequence {$x_n$} in ascending order: $x_1 \leq x_2 \leq \cdots \leq x_n$. Here, we can roughly observe the relationship between their magnitudes. It seems reasonable to assume $x_k = (-1)^{k-1}$. One intuitive understanding is that we need to balance the magnitudes on the left and right, ensuring they are neither too large nor too small. However, yesterday afternoon, @sky.mty provided a more rigorous explanation for me! He said that we consider the expansion of the square on the right-hand side: $(\sum\limits_{k=1}^na_kx_k)^2=\sum\limits_{k=1}^na_k^2x_k^2+2\sum\limits_{1\le i<j\le n}a_ia_jx_ix_j=\sum\limits_{k=1}^nx_k^2+2\sum\limits_{1\le i<j\le n}\{-1,1\}x_ix_j\le \sum\limits_{k=1}^na_kx_k^2$. We hope that $\sum\limits_{1\leq i<j\leq n} \{-1,1\} x_i x_j$ becomes smaller. Therefore, we want the coefficient of $x_ix_j$ to have as many -1 as possible. That's why we considered to take $a_k=(-1)^k$. But anyway, we both guessed this way this morning, so I think it's easy to think of it! Next, we will prove that $\sum\limits_{k=1}^n(-1)^{k-1}x_k^2\ge \left(\sum\limits_{k-1}^n(-1)^{k-1}x_k\right)^2$. Proof: We proceed by induction. The base case for $n=1,2,3$ are trivial. Now, assume that when $n$, the proposition is $Q_n$ .We suppose that $Q_1,Q_,\cdots ,Q_n$ are all right. Since we can't be sure the coefficient of the last term in the left-hand side sum, we consider proving it in the following way: $Q_{n-1}\rightarrow Q_{n+1},Q_n\rightarrow Q_{n+2}$. (1)If $n$ is odd. We have $\sum\limits_{k=1}^n(-1)^{k-1}x_k^2\ge \left(\sum\limits_{k-1}^n(-1)^{k-1}x_k\right)^2$. We prove that $\sum\limits_{k=1}^{n+2}(-1)^{k-1}x_k^2\ge \left(\sum\limits_{k-1}^{n+2}(-1)^{k-1}x_k\right)^2$. because $ \sum\limits_{k=1}^{n+2}(-1)^{k-1}x_k^2=\sum\limits_{k=1}^{n}(-1)^{k-1}x_k^2-x_{n+1}^2+x_{n+2}^2\ge \left(\sum\limits_{k-1}^n(-1)^{k-1}x_k\right)^2-x_{n+1}^2+x_{n+2}^2$. Which is equal to the case of $n=3$ (It's clear that $\sum\limits_{k=1}^n(-1)^{k-1}x_k\le x_{n+1}\le x_{n+2}$). Thus, the conclusion is established. (2)If $n$ is even. We have $ \sum\limits_{k=1}^{n+2}(-1)^{k-1}x_k^2=x_1^2-x_2^2+\sum\limits_{k=3}^{n+1}(-1)^{k-1}x_k^2\ge x_1^2-x_2^2+\left( \sum\limits_{k=3}^{n+2}(-1)^{k-1}x_k\right)^2$. Which we came back to the case of $n=3$ again (And it's also clear that $x_1\le x_2\le \sum\limits_{k=3}^{n+2}(-1)^{k-1}x_k $!) Thus, the conclusion is also established. Q.E.D