Flaw in A.M.-G.M. inequality for finding the range of function. Let $f(x) =2^x+2^ {-x} +3^x+3^ {-x} +4$. The minimum value of this function is 8.
 Which comes as
$\dfrac{2^x+\frac{1}{2^x}}{2}\ge \sqrt{2^x\cdot\frac{1}{2^x}}$
$2^x+2^{-x}\ge 2$
similarly,
$3^x+3^{-x}\ge 2$
Both these terms have minimum value at x=0
But what if we us the inequality as-
$\dfrac{2^x+2^{-x}+3^x+3^{-x}+4}{5}\ge\sqrt[5]4$
Hence the minimum value comes out to be $\sqrt[5]4$ which is less than 8.
I cannot see any flaw in both of the methods. What am I missing?

There's no guarantee there ever has to be situation where the LHS value has to match the RHS value when using the AM-GM inequality. The inequality only states the $2$ may be equal, but the LHS is always at least the size of the RHS.
In your last equation, note that since $2^x + 2^{-x} \ge 2$ and $3^x + 3^{-x} \ge 2$, then $\frac{2^x + 2^{-x} + 3^x + 3^{-x} + 4}{5} \ge \frac{2 + 2 + 4}{5} = \frac{8}{5} \gt \sqrt[5]{4}$.
