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.

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- Nov 29th 2009, 06:24 PMdannyboycurtisHolder'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. - Nov 30th 2009, 03:12 AMEnrique2
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. - Nov 30th 2009, 03:17 PMdannyboycurtis
Im not sure how to apply the concept of concavity to this problem...

- Nov 30th 2009, 10:39 PMredsoxfan325
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 $\displaystyle t\in[0,1]$.