I need to solve the following inequality using the MVT, not sure how to go about doing this:
$\displaystyle a^{t}b^{1-t}\leq ta+(1-t)b$ for $\displaystyle t\in [0,1], a>0,b>0$
Most of these problems are solved by finding the correct interval $\displaystyle [a,b]$ noting that $\displaystyle f(b)-f(a)=(b-a)f'(c)$ for some $\displaystyle c\in(a,b)$ and noticing that $\displaystyle f'(c)$ is either increasing or decreasing on $\displaystyle (a,b)$ so that $\displaystyle (b-a)f'(a)\le f(b)-f(a)\le f'(b)(b-a)$ or the decreasing analogue. Is thist one of those?