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Problem 833

AIME late

Algebra Difficulty 5.5 Prove it

Prove that for all integers $n \geq 2$ , we have: $\sum_{k=2}^{n} \frac{1}{\sqrt[k]{(2 k)!}} \geq \frac{n-1}{2 n+2}$ .
N.B. If $a>0$ , we denote $\sqrt[k]{a}$ the unique positive real number $b$ such that $b^{k}=a$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

We reason by induction on $n \geq 2$ .
For $n=2$ , we have $\frac{1}{\sqrt{24}} > \frac{1}{6}$ .
Suppose the desired inequality holds for the value $n-1 \geq 2$ . For the value $n$ , the right-hand side increases by $\frac{n-1}{2 n+2} - \frac{n-2}{2 n} = \frac{1}{n(n+1)}$ .
According to the induction hypothesis, it suffices to prove that $\frac{1}{\sqrt[n]{(2 n)!}} \geq \frac{1}{n(n+1)}$ .
For $k=1,2, \cdots, n$ , we have $(n-k)(n-k+1) \geq 0$ , so $0 \leq k(2 n-k+1) \leq n(n+1)$ .
By multiplying these inequalities term by term, we get $(2 n)! \leq (n(n+1))^{n}$ , from which the conclusion follows.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.