Given that $c$ satisfies $(\frac{c}{e})^c = e$, prove for $b \geq c$ that $b^b \geq e^{2b - c +1}$.

Question

Let $c$ be the number which solves the equation $(\frac{c}{e})^c = e$, note $c = 3.591...$. Show for $b \geq c$ we have $b^b \geq e^{2b - c +1}$.

This sentence showed up in a set of notes I am reading. My attempt began by observing that $c^c = e^{c+1}$ and so multiplying both sides by $e^{2b-2c}$ we get $e^{2b-c+1} \leq e^{2b-2c}c^c$. Hence we have to show that $e^{2b-2c}c^c \leq b^b$. Taking logarithms this is equivalent to showing that $b(\log b - 2) \geq c(\log c - 2)$. This clearly holds for $b \geq e^2$ but I'm worried about the $b < e^2$ case, and I'm not sure how to proceed.

Answer 1

Let $f(b)=b\ln b-2b$, then by rearranging the equation for $c$, we know that $f(c)=1-c$. \begin{align*} f'(b)=\ln b-1\geq 0\text{ for }b\geq e \end{align*} In particular, the function is increasing for $b\geq c$. So for all $b\geq c$, we have \begin{align*} b\ln b&\geq 2b-c+1\\ \implies b^b&\geq e^{2b-c+1} \end{align*}