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.
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.
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 $\displaystyle t\in[0,1]$.