Consider a positive integer $n$ and the real numbers $a_1 \leqslant a_2 \leqslant \cdots \leqslant a_n$ such that
$\displaystyle a_1 + 2a_2 + \cdots + na_n = 0$
Prove that
$\displaystyle a_1[x] + a_2[2x] + \cdots + a_n[nx] \geqslant 0$
for every $x$, where $[t]$ is the integer part of $t$
My attempt
Consider $M=\left \{ 1,2,...,n \right \}$ and $Y=\left \{ k\in M|a_k\geq 0 \right \},X=\left \{ k\in M|a_k<0 \right \}$ $\forall x\in\mathbb{R} \Rightarrow k[x]\leq [kx]\leq kx$
$\forall k\in Y \Rightarrow k[x]a_k\leq [kx]a_k$
$\forall k\in X\Rightarrow kxa_k\leq [kx]a_k$
$\Rightarrow \sum_{X} k[x]a_k+\sum_{Y}kxa_k\leq \sum_{k=1}^{n}a_k[kx]$
$\sum_{X} k[x]a_k+\sum_{Y}kxa_k=\sum_{X}a_k([x]-x)\geq 0$
So $\sum_{k=1}^{n}a_k[kx] \geq 0$
Is there a mistake in the above solution