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This question already has an answer here:

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?

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marked as duplicate by Macavity inequality Feb 2 at 11:40

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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 $(1,2,...,n)$.

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  • $\begingroup$ Always with the inequalities, eh? Congrats on getting $100$k rep, and a $(+1)$ from me :D $\endgroup$ – Mr Pie Feb 2 at 13:46

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