Use the Mean Value Theorem to prove the following inequalities:
1.$\ b^{b} -a^{a} > a^{a}(b-a)$ for all $a,b ∈ ℝ$ such that $a ≥ 1$ and $b ≥ 1$
2.$\frac{lnb}{b} - \frac{lna}{a} ≤ \frac{b-a}{a^2}$ such that $1≤ a < b ≤e$
I tried using the mean value theorem on both but I still don't understand why $a^a$ appears on the first case and $a^2$ appears on the second one.Any help would be kindly appreciated.
Let $f(x) = x^{x}$, then by MVT, $b^{b} - a^{a} = \xi^{\xi}(1 + ln\xi)(b - a)$, for some $\xi \in (a,b)$. Now, $x^{x}$ is strictly increasing in $[1,\infty)$ which implies $\xi^{\xi}(1 + ln\xi) > a^{a}$ and hence you obtain the desired inequality.
Let $f(x) = \frac{lnx}{x}$, then by MVT, $\frac{lnb}{b} - \frac{lna}{a} = \frac{1 - ln\xi}{\xi^{2}}(b-a)$, for some $\xi \in (a,b)$. Now observe that for $\xi \in (a,b) \subset [1,e]$, $\frac{1 - ln\xi}{\xi^{2}} \leq \frac{1}{\xi^{2}} \leq \frac{1}{a^{2}}$ and hence you obtain the desired inequality.