Minimum of the Sum of Logarithms

[anemone]

Find the minimum of $\large \log_{a_1}\left(a_2-\dfrac{1}{4}\right)+\log_{a_2}\left(a_3-\dfrac{1}{4}\right)+\cdots+\log_{a_n}\left(a_1-\dfrac{1}{4}\right)$ where $a_1,\,a_2,\cdots,a_n$ are real numbers in the interval $\left(\dfrac{1}{4},\,1\right)$.

 

[chisigma]

Find the minimum of $\large \log_{a_1}\left(a_2-\dfrac{1}{4}\right)+\log_{a_2}\left(a_3-\dfrac{1}{4}\right)+\cdots+\log_{a_n}\left(a_1-\dfrac{1}{4}\right)$ where $a_1,\,a_2,\cdots,a_n$ are real numbers in the interval $\left(\dfrac{1}{4},\,1\right)$.

[sp]Attempt ... because changing the order of the variables the problem remains the same, the solution will be such that $ a_ {1} = a_ {2} = ... = a_ {n} = x $ and the function to be minimized is...

$\displaystyle f(x) = log_{x} (x - \frac{1}{4})\ (1)$

Proceeding in standard fashion is...

$\displaystyle f^{\ '} (x) = \frac{\frac{\ln x}{x - \frac{1}{4}} - \frac{\ln (x - \frac{1}{4})}{x}}{\ln ^{2} x}\ (2)$

... and the (2) vanishes for...

$\displaystyle x\ \ln x = (x - \frac{1}{4})\ \ln (x - \frac{1}{4}) \implies x = \frac{1}{2}\ (3)$[/sp]

Kind regards

$\chi$ $\sigma$

 

[anemone]

[sp]Attempt ... because changing the order of the variables the problem remains the same, the solution will be such that $ a_ {1} = a_ {2} = ... = a_ {n} = x $ and the function to be minimized is...

$\displaystyle f(x) = log_{x} (x - \frac{1}{4})\ (1)$

Proceeding in standard fashion is...

$\displaystyle f^{\ '} (x) = \frac{\frac{\ln x}{x - \frac{1}{4}} - \frac{\ln (x - \frac{1}{4})}{x}}{\ln ^{2} x}\ (2)$

... and the (2) vanishes for...

$\displaystyle x\ \ln x = (x - \frac{1}{4})\ \ln (x - \frac{1}{4}) \implies x = \frac{1}{2}\ (3)$[/sp]

Kind regards

$\chi$ $\sigma$

Thanks chisigma for participating in this challenge!

Spoiler

And in your method,the minimum of $\large \log_{a_1}\left(a_2-\dfrac{1}{4}\right)+\log_{a_2}\left(a_3-\dfrac{1}{4}\right)+\cdots+\log_{a_n}\left(a_1-\dfrac{1}{4}\right)$ where $a_1,\,a_2,\cdots,a_n$ are real numbers in the interval $\left(\dfrac{1}{4},\,1\right)$ would be $2n$.

Here is the solution of other that I wanted to share:

Spoiler

Since $\log_m a$ is a decreasing function of $a$ when $0<m<1$, and since $\left(a-\dfrac{1}{2}\right)^2\ge 0$ which implies $a^2\ge a-\dfrac{1}{4}$, we have

$\large \log_{a_k}\left(a_{k+1}-\dfrac{1}{4}\right)\ge \log_{a_k} a_{k+1}^2=2\log_{a_k} a_{k+1}=2\dfrac{\log a_{k+1}}{\log a_{k}}$

It follows that

$\begin{align*}\log_{a_1}\left(a_2-\dfrac{1}{4}\right)+\log_{a_2}\left(a_3-\dfrac{1}{4}\right)+\cdots+\log_{a_n}\left(a_1-\dfrac{1}{4}\right)&\ge 2\left(\dfrac{\log a_2}{\log a_1}+\dfrac{\log a_3}{\log a_2}+\cdots+\dfrac{\log a_n}{\log a_{n-1}}+\dfrac{\log a_1}{\log a_n}\right)\\&\ge 2n\,\,\,\text{by AM-GM inequality}\end{align*}$

Equalities hold iff $a_1=a_2=\cdots=a_n=\dfrac{1}{2}$.