Prove $\forall t\in [0,1):\, t\le \frac{1-t^t}{1-t}$ How do I prove $$\forall t\in [0,1):\,t\le \frac{1-t^t}{1-t}?$$
Do not use derivatives or integrals and assume that irrational exponentiation is defined by limits and define $0^0=1$.
My attempt:
Let $t=\frac{1}{a}$, thus $a\gt 1$. The case for $t=0$ is trivial. So
$$\begin{align}\frac{1}{a}&\le \frac{1-\left(\frac{1}{a}\right)^{\frac{1}{a}}}{1-\frac{1}{a}}\\&=\frac{\left(1-a^{-\frac{1}{a}}\right)a}{a-1}\\&=\frac{a-a^{1-\frac{1}{a}}}{a-1}\\a&\ge \frac{a-1}{a-a^{1-\frac{1}{a}}}\\a^2-a^{2-\frac{1}{a}}-a+1&\ge 0.\end{align}$$
Now $a^{2-\frac{1}{a}}\le a^2$ but I don't know how to use this fact to compare $a^{2-\frac{1}{a}}+a$ to $a^2$.
 A: 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
A: 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.
A: 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.
A: Since $\;t\ln t<0\;$ for all $\;t\in\left]0,1\right[,\;$ it results that
$t^t-1=e^{t\ln t}-1<t\ln t+\frac{1}{2}t^2\ln^2 t\;\;$ for all $\;t\in\left]0,1\right[$.
Therefore, $$t-\frac{1-t^t}{1-t}=\frac{t-t^2-1+t^t}{1-t}<\frac{t-t^2+t\ln t+\frac{1}{2}t^2\ln^2 t}{1-t}=\\=\frac{t}{1-t}\left(1-t+\ln t+\frac{1}{2}t\ln^2t\right)\;\;\text{ for all }\;t\in\left]0,1\right[.\color{blue}{\quad(*)}$$
Let $\;\phi(t):\left]0,1\right]\to\mathbb{R}\;$ be the function defined as
$\phi(t)=1-t+\ln t+\frac{1}{2}t\ln^2 t$.
$\phi(t)$ is differentiable on $\left]0,1\right]$ and
$\phi’(t)=-1+\frac{1}{t}+\frac{1}{2}\ln^2 t+\ln t\ge-1+\frac{1}{t}+\ln t\;\;$ for all $\;t\in\left]0,1\right].$
Since $\;\ln(1+x)<x\;\;\forall x\in\left]-1,+\infty\right[\setminus\left\{0\right\}\;$ and $\;-1+\frac{1}{t}>0\;\;\forall t\in\left]0,1\right[,\;$ it results that
$-\ln t=\ln\left(1-1+\frac{1}{t}\right)<-1+\frac{1}{t}\;\;$ for all $\;t\in\left]0,1\right[$.
Hence $\;-1+\frac{1}{t}+\ln t>0\;\;$ for all $\;t\in\left]0,1\right[$.
So $\;\phi’(t)>0\;\;$ for all $\;t\in\left]0,1\right[$.
Therefore $\;\phi(t)$ is an increasing function on $\left]0,1\right]\;$ and
$1-t+\ln t+\frac{1}{2}t\ln^2 t=\phi(t)<\phi(1)=0\;\;$ for all $\;t\in\left]0,1\right[$.
Since $\;1-t+\ln t+\frac{1}{2}t\ln^2 t<0\;\;$ and $\;\;\frac{t}{1-t}>0\;\;$ for all $\;t\in\left]0,1\right[,\;$ from $(*)$ it follows that
$$t<\frac{1-t^t}{1-t}$$ for all $\;t\in\left]0,1\right[$.
A: The generalised inequality can be proven as follows: Let $$f:[0,1]\to[0,1-r],\\t\mapsto1-r^t$$ for $r\in]0,1[$. Since $$r^t=\exp(t\ln(r))$$ is convex for $t\in[0,1]$ (you can prove this "with or without" derivatives), $f$ is concave and hence $$t(1-r)=(1-t) f(0)+t f(1)\le f(t)=1-r^t$$ and we are done.
The inequality in your question is the case $r=t$.
