Holder's Inequality

**dannyboycurtis** · November 29th 2009, 06:24 PM

I need to prove that the following inequality is true only if t=0, t=1 or a=b>0:
$a^{t}b^{1-t}\leq ta+(1-t)b$
I have no idea how to proceed.

**Enrique2** · November 30th 2009, 03:12 AM

The inequality is true for each 1<t<1 also! it is just the main ingredient for proving Holder inequality. Since log is incresing, the inequality is equivalent to
$log(a^tb^{1-t})\leq log(ta+(1-t)b)$

Now apply basic properties of the logarithm to the left hand side and use that log is concave to conclude.

**dannyboycurtis** · November 30th 2009, 03:17 PM

Im not sure how to apply the concept of concavity to this problem...

**redsoxfan325** · November 30th 2009, 10:39 PM

Originally Posted by dannyboycurtis

Im not sure how to apply the concept of concavity to this problem...

The concavity of log means that given two points $a$ and $b$ and the line $\ell(t)$ going through $\log(a)$ and $\log(b)$ , for any point $c\in[a,b]$ , $\log(c)\geq\ell(c)$ .

Represent the point $c$ as $c=ta+(1-t)b$ for some $t\in[0,1]$ .

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