1. ## Holder's Inequality

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

2. 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
$\displaystyle log(a^tb^{1-t})\leq log(ta+(1-t)b)$

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

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

4. 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 $\displaystyle a$ and $\displaystyle b$ and the line $\displaystyle \ell(t)$ going through $\displaystyle \log(a)$ and $\displaystyle \log(b)$, for any point $\displaystyle c\in[a,b]$, $\displaystyle \log(c)\geq\ell(c)$.

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