Mathematical Reflections 2012, Issue 6 - Problem S249

Problem:
Find the minimum of $2^x-4^x+6^x-8^x-9^x+12^x$ where $x$ is a positive real number.

Proposed by Titu Andreescu.

Solution:
We have $$\begin{array}{rcl}2^x-4^x+6^x-8^x-9^x+12^x&=&(4^x-3^x)(3^x-2^x)-[(4^x-3^x)+(3^x-2^x)]\\&=&(4^x-3^x-1)(3^x-2^x-1)-1. \end{array}$$
Let $f_n(x)=(n+1)^x-n^x-1$, where $n$ is a positive integer. $f_n(x)$ is increasing since
$$f'_n(x)=(n+1)^x\log(n+1)-n^x\log n > 0 \iff \left(1+\dfrac{1}{n}\right)^x>\dfrac{\log n}{\log (n+1)}$$ for all $x \in \mathbb{R}^+$. Moreover $f_n(1)=0$, so $f_n(x) \geq 0$ if $x \in [1,\infty)$ and $f_n(x) \leq 0$ if $x \in (0,1]$, for all $n \in \mathbb{N}^*$. Therefore, $$f_3(x)f_2(x)-1 \geq -1 \qquad \forall x \in \mathbb{R}^+$$ and the equality occurs if and only if $x=1$.