We can conduct the following moves to a real number $x$: choose a positive integer $n$, and positives reals $a_1,a_2,\cdots, a_n$ whose reciprocals sum up to $1$. Let $x_0=x$, and $x_k=\sqrt{x_{k-1}a_k}$ for all $1\leq k\leq n$. Finally, let $y=x_n$. We said $M>0$ is "tremendous" if for any $x\in \mathbb{R}^+$, we can always choose $n,a_1,a_2,\cdots, a_n$ to make the resulting $y$ smaller than $M$. Find all tremendous numbers. Proposed by ckliao914.
Problem
Source: 2023 IMOC A5
Tags: IMOC, algebra, inequalities
Sourorange
03.12.2023 11:55
Any solutions? The problem seems to be a bit hard to guess the answer.
lkjin7932
19.12.2023 18:12
Solution
Taking logarithms and using Jensen inequality easily leads to the result.
Proof is just calculation
zhihanpeng
23.08.2024 09:18
We can obtain \[ \begin{aligned} & y = x^{\frac{1}{2^n}} a_1^{\frac{1}{2^n}} \ldots a_{n-1}^{\frac{1}{2^2}} a_n^{\frac{1}{2}} \\ & = \left(x a_1 a_2^2 \cdots a_{n-1}^{2^{n-2}} a_n^{2^{n-1}}\right)^{\frac{1}{2^n}} \\ & = \left[x\left(\left(\frac{1}{a_1}\right)\left(\frac{1}{a_2}\right)^2 \cdots \left(\frac{1}{a_n}\right)^{2^{n-1}}\right)^{-1}\right]^{\frac{1}{2^n}} \\ & = \left[x \cdot \left(\frac{1}{2^{n-1 + (n-2) \cdot 2^1 + \cdots + 1 \cdot 2^{n-2}}}\left(\frac{2^{n-1}}{a_1}\right)\left(\frac{2^{n-2}}{a_2}\right) \left(\frac{2^{n-2}}{a_2}\right) \cdots \left(\frac{1}{a_n}\right) \cdots \left(\frac{1}{a_n}\right)\right)^{-1}\right]^{\frac{1}{2^n}} \\ & \geq \left(x \cdot \frac{2^{2^n - n - 1}}{\left(\frac{2^{n-1}}{2^n - 1}\right)^{2^n-1}}\right)^{\frac{1}{2^n}} \\ & = \left(\frac{x \cdot \left(2^n - 1\right)^{2^n - 1}}{2^{(n-2) \cdot 2^n + 2}}\right)^{\frac{1}{2^n}} = f(n, x) \end{aligned} \] We need to find the best upper bound for all values of \[ g(x) = \min_{n \geq 1} f(n, x). \] \[ f(n, x) < \left(\frac{x \cdot \left(2^n\right)^{2^n - 1}}{2^{(n-1) \cdot 2^n + 2}}\right)^{\frac{1}{2^n}} = \left(\frac{x}{2^{n + 2 - 2^{n+1}}}\right)^{\frac{1}{2^n}} = \frac{x^{\frac{1}{2^n}}}{2^{\frac{n+2}{2^n} - 2}} = \frac{4x^{\frac{1}{2^n}}}{2^{\frac{n+2}{2^n}}} \] \[ < 4 \cdot \left(\frac{x}{2^n}\right)^{\frac{1}{2^n}}. \] Choose large \(n \Rightarrow f(n, x) < 4 \Rightarrow g(x) < 4\). Let \(x \rightarrow +\infty\). Then because the optimal \(n\) must satisfy \(2^n > x\), we also have \(n \rightarrow +\infty\). Then \(\left(\frac{2^n}{2^n - 1}\right)^{\frac{2^n - 1}{2n}} \rightarrow 1\), \(\frac{2^{\frac{n+2}{2^n}}}{2^{\frac{n}{2^n}}} \rightarrow 1\). So \(f(n, x) \rightarrow 4 \cdot \left(\frac{x}{2^n}\right)^{\frac{1}{2n}}\). Consider \(f(y) = \left(\frac{x}{y}\right)^{\frac{1}{y}}\). \(f^{\prime}(y) = 0 \Rightarrow y = e x\). \(f(y) \geq \left(\frac{1}{e}\right)^{\frac{1}{ex}} \rightarrow 1\), when \(x \rightarrow +\infty\). Thus, we have \(M = 4\) as the optimal bound.