Inequality $\frac{a_1}{1^2}+\frac{a_2}{2^2}+...+\frac{a_n}{n^2}\ge\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}$ Suppose $a_i$ are dinstinct positive integers $\forall1\le i\le n$. Prove that
$$\frac{a_1}{1^2}+\frac{a_2}{2^2}+...+\frac{a_n}{n^2}\ge\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}$$

My approach:
I will procede by proof by contradiction. If $a_1,a_2,...,a_n$ is not a permutation of $1,2,...,n$, then one can reduce $a_k (\exists a_k>n)$ to a value $<n$ and $>1$ So that $a_1,a_2,...,a_n$ is a permutation of $1,2,...,n$. From now on assume $a_1,a_2,...,a_n$ is a permutation of $1,2,...,n$. Now, by the rearrangement inequality, $$\frac{a_1}{1^2}+\frac{a_2}{2^2}+...+\frac{a_n}{n^2}\ge\frac{1}{1^2}+\frac{2}{2^2}+...+\frac{n}{n^2}=\text{LHS}$$ with the rearrangement sequence being $\{a_1,a_2,...,a_n\} \text{and} \{\frac{1}{1^2},\frac{1}{2^2},...,\frac{1}{n^2}\}$.

Is this proof correct? Is there a better way of doing it?

I think your proof is correct, but I think it's better to get it without contradiction.
$(1,2,...,n)$ and $\left(\frac{1}{1^2},\frac{1}{2^2},...,\frac{1}{n^2}\right)$ have an opposite ordering. 
Thus, by Rearrangement
$$\frac{a_1}{1^1}+\frac{a_2}{2^2}+...+\frac{a_n}{n^2}\geq\frac{a'_1}{1^1}+\frac{a'_2}{2^2}+...+\frac{a'_n}{n^2}\geq$$
$$\geq\frac{1}{1^1}+\frac{2}{2^2}+...+\frac{n}{n^2}=1+\frac{1}{2}+...+\frac{1}{n}.$$
Here $(a_1',a_2',...,a_n')$ is a permutation of $(1,2,...,n)$.
