The infimum of the given set

Question

What is the infimum of $A = \{3^{2x} + 3^{\frac{1}{2x}} | x > 0\}$?

Without guessing, how do you determine the infimum of that set? Many advises tell us to guessing $x$, but for this problem, it is not sound that easy. Please, help me to solve this.

Answer 1

By Arithmetic-Geometric inequality, \begin{align*} 3^{2x}+3^{1/(2x)}\geq 2(3^{2x}\cdot3^{1/(2x)})^{1/2}=2\cdot 3^{(2x+(2x)^{-1})/2}\geq 2\cdot 3^{((2x)(2x)^{-1})^{1/2}}=2\cdot 3=6, \end{align*} and the equality holds if and only if $3^{2x}=3^{1/(2x)}$ and $2x=(2x)^{-1}$. For $x>0$, we can take $x=1/2$ to satisfy the condition.

Answer 2

One term is strictly increasing, the other strictly decreasing on $(0,\infty)$ so the minimum (equal to infimum due to continuoty) is when their exponents are equal, namely $x=\frac{1}{2},$ giving infimum 6.