# Holder's Inequality

• Nov 29th 2009, 07:24 PM
dannyboycurtis
Holder's Inequality
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.
• Nov 30th 2009, 04:12 AM
Enrique2
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.
• Nov 30th 2009, 04:17 PM
dannyboycurtis
Im not sure how to apply the concept of concavity to this problem...
• Nov 30th 2009, 11:39 PM
redsoxfan325
Quote:

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 $t\in[0,1]$.