Proving Inequality

Solving for the critical points of the LHS as a function of $a$ gives \begin{align*} x\left(1+\frac{a}{b}\right)^{x-1}\frac{1}{b} + x\left(1+\frac{b}{a}\right)^{x-1}(-ba^{-2}) &= 0\\ a^{x+1}(b+a)^{x-1}-b^{x+1}(a+b)^{x-1}&=0\\ a&=b, \end{align*} since all variables are positive. Clearly the LHS diverges as $a\to 0$ or $a\to\infty$, so this critical point is the global minimum and the inequality follows.

This can be solved by AM-GM for 2 variables, i.e. $m+n\ge 2\sqrt{mn}$. Divide by $2^x$, let $t=\frac ab$ and this turns to $$\left(\frac{1+t}{2}\right)^{x} + \left(\frac{1+\frac{1}{t}}{2}\right)^{x} \ge 2.$$

Applying the AM-GM on the LHS, you get: $$\left(\frac{1+t}{2}\right)^{x} + \left(\frac{1+\frac{1}{t}}{2}\right)^{x} \ge 2\left(\frac{(1+t)(1+\frac{1}{t})}{4}\right)^{x/2}$$

It remains to show $(1+t)\Bigl(1+\frac{1}{t}\Bigr)\ge 4$. This can be proved by Hölder in the case of $p=q=\frac12, n=2$ (actually, AM-GM for 2 variables can be also proved by Hölder).