Prove that : $\log_{n}(2)\log_{n}(4)\log_{n}(6)...\log_{n}(2n-2)\leq 1$ for all $2≤n\in \mathbb N$

Question

Problem :

Let $n$ be a natural number then for all $n\geq 2$ prove the following inequality :

$\log_{n}(2)\log_{n}(4)\log_{n}(6)...\log_{n}(2n-2)\leq 1$

I see the help say use :

$\sqrt ab\leq \frac{a+b}{2}$

$\log_{n}(2)\log_{n}(4)≤\frac{\ln 4+\ln 2}{2\ln n}$

But I don't know how I use this help ?

I would like see other method

Answer 1

As has been noted, $\log_nx=\frac{\ln x}{\ln n}$. The trick is to first rearrange the desired result to$$\frac{1}{n-1}\sum_{k=1}^{n-1}\ln\ln(2k)\le\ln\ln n.$$Notice $n-1$ the arguments $2,\,\cdots,\,2n-2$ of $\ln\ln x$ on the left-hand side have arithmeti mean $n$. So we can then use Jensen's inequality for concave functions: $\ln\ln x$ has second derivative $-\frac{1+\ln x}{x^2\ln^2x}$, which is negative for $x>\frac1e$.

Answer 2

You can use GM-AM and the concavity of $\log_n(x)$ (Jensen's inequality - concave case)

\begin{eqnarray*}\sqrt[n]{1\cdot\prod_{k=1}^{n-1}\log_n 2k} & \stackrel{GM-AM}{\leq} & \frac{\overbrace{1}^{=\log_n n}+\sum_{k=1}^{n-1}\log_n 2k}{n} \\ & \stackrel{\log_n \; concave}{\leq} & \log_n\left(\frac{n+\sum_{k=1}^{n-1}2k }{n}\right) \\ & = & \log_n\left(1+ (n-1)\right) = 1 \end{eqnarray*}