Prove $\forall t\in [0,1):\, t\le \frac{1-t^t}{1-t}$

Take the change of variables, $t = 1- x$, then rearranging, the inequality becomes $$ (1- x)^{ 1-x } \leq x(x - 1) + 1 $$ which is true by Bernoulli's Inequality

Let $s$ be any real number in $\left]0,1\right[$ and prove that $\left\{a_n\right\}_{n\in\mathbb{N}}=\left\{\frac{1-s^n}{n}\right\}_{n\in\mathbb{N}}$ is a decreasing sequence.

Since $\;s\in\left]0,1\right[,\;$it results that

$ns^n<1+s+s^2+\ldots+s^{n-1}\;\;$ for all $\;n\in\mathbb{N}.$

Hence,

$\frac{s^n}{1+s+s^2+\ldots+s^{n-1}}<\frac{1}{n}\;\;$ for all $\;n\in\mathbb{N}$,

$\frac{1+s+s^2+\ldots+s^{n-1}+s^n}{1+s+s^2+\ldots+s^{n-1}}<1+\frac{1}{n}\;\;$ for all $\;n\in\mathbb{N}$,

$\frac{(1-s)(1+s+s^2+\ldots+s^{n-1}+s^n)}{(1-s)(1+s+s^2+\ldots+s^{n-1})}<\frac{n+1}{n}\;\;$ for all $\;n\in\mathbb{N}$,

$\frac{1-s^{n+1}}{1-s^n}<\frac{n+1}{n}\;\;$ for all $\;n\in\mathbb{N}$,

$\frac{1-s^{n+1}}{n+1}<\frac{1-s^n}{n}\;\;$ for all $\;n\in\mathbb{N}$,

$a_{n+1}<a_n\;\;$ for all $\;n\in\mathbb{N}$.

So the sequence $\left\{a_n\right\}_{n\in\mathbb{N}}=\left\{\frac{1-s^n}{n}\right\}_{n\in\mathbb{N}}$ is monotonically decreasing for all $s\in\left]0,1\right[$.

Let $\;r\;$ be any real number in $\left]0,1\right[$ and let $\;p, q\in\mathbb{N}\;$ such that $\;p<q$.

If $\;s=r^{\frac{1}{q}}$ then $s\in\left]0,1\right[$ and, since $\left\{a_n\right\}_{n\in\mathbb{N}}$ is decreasing, we get that

$\frac{1-r^{\frac{p}{q}}}{1-r}=\frac{1-s^p}{1-s^q}=\frac{p\cdot a_p}{q\cdot a_q}>\frac{p}{q}.$

So we have proved that

$\frac{1-r^t}{1-r}>t\;\;$ for all $\;r\in\left]0,1\right[\;$ and for all $\;t\in\left]0,1\right[\cap\mathbb{Q}$.

By continuity of the function $\;f(t)=\frac{1-r^t}{1-r}-t\;$ on $\left]0,1\right[$, we also get that

$\frac{1-r^t}{1-r}\ge t\;\;$ for all $\;r\in\left]0,1\right[\;$ and for all $\;t\in\left]0,1\right[.$

I have proved it without using AM-GM inequality or Bernoulli’s inequality or concavity. I only used continuity.

Assume that we know: $\quad t^t$ is continuous on $(0, 1)$.

It suffices to prove that $$t^t \le 1 + t(t-1), \ 0 < t < 1. \tag{1}$$

First, (1) is true for rational $t\in (0, 1)$. Indeed, let $t = \frac{m}{n}$ with $0 < m < n$. By AM-GM, we have $$\sqrt[n]{t^m} \le \frac{1\cdot (n-m) + t \cdot m}{n} = 1 + \frac{m}{n}(t-1) = 1 + t(t-1).$$

Second, suppose $r^r > 1 + r(r-1)$ for some irrational $r\in (0, 1)$. By continuity, there exists $a < r < b$ such that $x^x > 1 + x(x-1)$ for all $x$ in $(a, b)$. Contradiction.

We are done.