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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 two different Hölder's Inequality. It seems the one with $1/p+1/q=1$ definitely does not suit here. For different inequalityes see the link below.

Added : All variables above are positive.

P.S. For Inequalities Follow This.

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for positive a and b? –  Bananarama Jan 3 '13 at 23:55
    
Yes! That is given! –  007resu Jan 3 '13 at 23:56
    
@Alex, that doesn't make sense if $x$ is not an integer, which it most likely need not be. Also, not true, unless you have a +1 or -1 somewhere. –  Calvin Lin Jan 4 '13 at 1:24
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2 Answers

up vote 3 down vote accepted

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).

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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.

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This doesn't use Holder's Inequality though. –  Alex R. Jan 4 '13 at 0:13
    
@Alex, right. I am rather looking solution that uses Hölder's inequality. –  007resu Jan 4 '13 at 0:26
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@Alex, user, fair enough. I'll leave this here in case any solution is more useful than none. –  user7530 Jan 4 '13 at 0:30
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