# Inequality $\frac{a_1}{1^2}+\frac{a_2}{2^2}+…+\frac{a_n}{n^2}\ge\frac{1}{1}+\frac{1}{2}+…+\frac{1}{n}$ [duplicate]

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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 $$ and $$>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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• Proof is correct. I can't think of a better way to do this. – Neo Darwin Feb 2 at 8:55
• – Martin R Feb 2 at 9:45

## 1 Answer

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

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