Let f:RR be such that for all xR, (2^1+x+2^1-x), f(x), (3^x+3^-x) are in A.P.

Question

Let $f:\mathbb{R}	o\mathbb{R}$ be such that for all $x\in\mathbb{R}$, $\left(2^{1+x}+2^{1-x}ight), f(x), \left(3^x+3^{-x}ight)$ are in A.P. Then the minimum value of $f(x)$ is:

Accepted Answer

$$f(x) = \frac{\left(2^{1+x}+2^{1-x}\right)+\left(3^x+3^{-x}\right)}{2} = \left(2^x+\frac{1}{2^x}\right)+\left(\frac{3^x+3^{-x}}{2}\right)\cdots$$

By AM--GM, $2^x+2^{1-x} \ge 2\sqrt2$ and $3^x+3^{-x}\ge2$; combining terms via AM--GM on each pair gives $f(x)\ge3$.

Solutions · 1