Let $A$ the first sum, $B$ the second and $C$ the third. from cauchy, we have $AC \ge B^2$. but, $AC = B^2$. the equality in cauchy occur if the terms are proportional. so, $jx_j/j^5x_j = ix_i/i^5x_i$ for all $i$,$j$ $\Longrightarrow i = j$, implossible. E.L

i believe the original question is non-negative reals.. also, i think there are six solns but some sources say there are only five.. n=0,1,4,9,16,25

Yes, the original question is.. "non-negative reals"... I just added a solution on the main page

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$a^2\sum_{k=1}^{5}{kx\textsubscript{k}}-2a\sum_{k=1}^{5}{k^3x\textsubscript{k}}+\sum_{k=1}^{5}{k^5x\textsubscript{k}}=a^3-2a^3+a^3$ $\sum_{k=1}^{5}{(a^2kx\textsubscript{k}-2ak^3x\textsubscript{k}+k^5x\textsubscript{k})}=0$ $\sum_{k=1}^{5}{k(a^2-2ak^2+(k^2)^2)x\textsubscript{k}}=0$ $\sum_{k=1}^{5}{k(a-k^2)^2x\textsubscript{k}}=0$ Because $x\textsubscript{k}\gneq0$: * all $x\textsubscript{k}=0$ --> $a=0$ * one $x\textsubscript{k}\neq0$ and the others $x\textsubscript{i}=0$ --> $a-k^2=0 \iff a=k^2$ --> $a=\{0,1,2^2,3^2,4^2,5^2\}$

C-S directly gives: $$\left(\sum_{k=1}^5kx_k\right)\left(\sum_{k=1}^5k^5x_k\right)^2\ge\left(\sum_{k=1}^5k^3x_k\right)^2.$$But both sides are actually equal, so by analyzing the equality case of C-S, we have $k^5x_k=kx_k\cdot c$ for some constant $c$ and all $k\in\{1,\ldots,5\}$. Then $c=k^4$ for each $k\in\{1,\ldots,5\}$ which is absurd, so no such sequence exists.