Prove the following
Question:

Let $f: R \rightarrow R$ be such that for all $x \in R,\left(2^{1+x}+2^{1-x}\right), f(x)$ and $\left(3^{x}+3^{-x}\right)$ are in A.P., then the minimum value of $f(x)$ is:__________.

Solution:

If $2^{1-x}+2^{1+x}, f(x), 3^{x}+3^{-x}$ are in A.P., then

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

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

Using AM $\geq$ GM

$f(x) \geq 3$