Let $n\geq 2$ be an integer and $a_1,a_2,\cdots,a_n>1$ be real numbers. Prove that the inequality below holds. $$\prod_{i=1}^n\left(a_ia_{i+1}-\frac{1}{a_ia_{i+1}}\right)\geq 2^n\prod_{i=1}^n\left(a_i-\frac{1}{a_i}\right)$$
Problem
Source: 2024 Turkey Junior National Olympiad P4
Tags: algebra, Inequality, inequalities, inequalities proposed
21.12.2024 00:02
Follows from $(xy-\frac{1}{xy})^2 > 4 (x-\frac{1}{x})(y-\frac{1}{y})$ $(xy-\frac{1}{xy})^2- 4 (x-\frac{1}{x})(y-\frac{1}{y})=4 (\frac{x}{y}+\frac{y}{x}-2)+\frac{(xy-1)^4}{x^2y^2} >0$
21.12.2024 05:56
Let $n\geq 2$ be an integer and $a_1,a_2,\cdots,a_n\geq 1$ be real numbers. Prove that the inequality below holds. $$\prod_{i=1}^n\left(1+a_ia_{i+1}-\frac{1}{a_ia_{i+1}}\right)\geq \prod_{i=1}^n\left(1+a_i-\frac{1}{a_i}\right)$$
21.12.2024 17:53
My solution during the exam. Define $b_i=a_i-\frac{1}{a_i}$. From AM-GM, $$a_ia_{i+1}+\frac{1}{a_ia_{i+1}}=b_ib_{i+1}+\frac{a_i}{a_{i+1}}+\frac{a_{i+1}}{a_i}\geq b_ib_{i+1}+2$$$$\Leftrightarrow \left(\sqrt{a_ia_{i+1}}-\frac{1}{\sqrt{a_ia_{i+1}}}\right)^2\geq b_ib_{i+1}$$holds. This is true for $i=1,2,\dots , n$. If we multiply for all values and take the square root we obtain $$\prod_{i=1}^n \left(\sqrt{a_ia_{i+1}}-\frac{1}{\sqrt{a_ia_{i+1}}}\right)\geq \prod_{i=1}^n (b_i)=\prod_{i=1}^n \left(a_i-\frac{1}{a_i}\right).$$This means that if $$a_ia_{i+1}-\frac{1}{a_ia_{i+1}}\geq 2\sqrt{a_ia_{i+1}}-\frac{2}{\sqrt{a_ia_{i+1}}}\Leftrightarrow (\sqrt{a_ia_{i+1}}-1)^2\geq 0$$is correct, the statement is proven. However this is always correct. $\blacksquare$ Note: Due to the given conditions, there is no equality case. If we pick $a_i=1+\varepsilon$ for all $i$, there is an equality case when $\varepsilon\rightarrow 0$.
21.12.2024 18:22
I hope there is no error.