Proving Inequality I stumbled upon the following inequality and I need to prove it. 
$$\left(1+\frac{a}{b}\right)^x+\left(1+\frac{b}{a}\right)^x\ge 2^{x+1}$$
I am expected to use Holder's Inequality but there seem to be two different Hölder's Inequality. It looks like the one with $1/p+1/q=1$ definitely does not suit here. For different inequalities see the link below.
Added : All variables above are positive.
P.S. For Inequalities Follow This.
 A: 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.
A: 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).
A: $$\underbrace{\left(1+\frac{a}{b}\right)^x}_\text{$2AM \geq 2GM $}
+\underbrace{\left(1+\frac{b}{a}\right)^x}_\text{$2AM \geq 2GM $} 
≥ \underbrace{\left(2\frac{a^{1/2}}{b^{1/2}}\right)^x+\left(2\frac{b^{1/2}}{a^{1/2}}\right)^x}_\text{$2AM \geq 2GM $}$$
$$ ≥2 \sqrt{\left(2\frac{b^{1/2}}{a^{1/2}}\right)^x\cdot\left(2\frac{a^{1/2}}{b^{1/2}}\right)^x}=2\sqrt{2^{2x}}=2^{x+1}$$
