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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 $a^2$.

Poder Rac
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5 Answers5

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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

cha21
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  • The question was motivated by $\forall t\in [0,1],, \forall r\in (0,1):, t\le \frac{1-r^t}{1-r}$. Do you think this method could be adapted to prove the general case with two variables? – Poder Rac Aug 02 '20 at 22:01
  • Yes, the proof is the exact same. – cha21 Aug 02 '20 at 22:05
  • This generalization of Bernoulli’s inequality can be proved by comparing derivatives. So you are using derivatives implicitly. You only showed that the inequality of the question is equivalent to a generalization of Bernoulli’s inequality, but if we proved it, we need to use derivatives. So I cannot understand why your answer was accepted and mine not. – Angelo Aug 02 '20 at 22:11
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    @Angelo You can prove the generalized Bernoulli's inequality by weighted AM–GM inequality. That makes use of the fact that $\ln$ is concave which doesn't need derivatives to be proved. – Poder Rac Aug 02 '20 at 22:12
  • So please show a proof that $\ln x$ is concave without using derivatives and without using AM-GM. I think that in any case derivatives are used implicitly. So I cannot understand why a answer is accepted only because it does not show derivatives explicitly. – Angelo Aug 02 '20 at 22:36
  • $\text{GM}\leq \text{AM}$ can be proven over rationals without recourse to logarithms or any analytic machinery. Leap-forward fallback induction plus sufficient 'padding' will do (chp 2 of Steele's CS Masterclass is a nice reference). It suffices to prove the OPs result then for $x\in(0,1)\cap \mathbb Q$ by a density plus continuous mapping argument. – user8675309 Aug 02 '20 at 22:40
  • @Angelo Here is a proof without derivatives: https://math.stackexchange.com/q/702241/144766 Exponentiation is defined via Taylor series. – mechanodroid Aug 02 '20 at 22:40
  • But coefficients of Taylor series contain derivatives. – Angelo Aug 02 '20 at 22:44
  • @Angelo If $\ln :\mathbb{R}^+\to\mathbb{R}$ and $\forall x_1,x_2\in\mathbb{R}^+:,\frac{\ln x_1+\ln x_2}{2}\le \ln \frac{x_1+x_2}{2}$, then $\ln$ is concave. The inequality is equivalent to $\sqrt{x_1 x_2}\le\frac{x_1+x_2}{2}$ after exponentiating. Since $0\le (x_1-x_2)^2=(x_1+x_2)^2-4x_1 x_2$, the proof is complete. – Poder Rac Aug 02 '20 at 23:27
  • Actually you have to prove that $t\ln x_1+(1-t)\ln x_2\le \ln\left(tx_1+(1-t)x_2\right)$ for all $t\in\left[0,1\right]$ and for all $x_1 , x_2\in\mathbb{R}^+$. So you need weighted AM-GM inequality. – Angelo Aug 03 '20 at 00:00
  • The property “midpoint concave $\implies$ concave” which is valid for continuous functions, needs a proof. – Angelo Aug 03 '20 at 00:27
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    @Angelo As you can see here, you actually don't need weighted AM–GM inequality to prove midpoint concave $\implies$ concave: https://math.stackexchange.com/questions/1002248/if-f-is-continuous-and-f-big-frac12xy-big-le-frac12-big-fx – Poder Rac Aug 03 '20 at 01:04
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    Ok, it means that by using induction, limits and continuity we can prove the inequality that there is in the question. We do not need derivatives but we need limits of sequences and continuity. – Angelo Aug 03 '20 at 01:31
  • @Angelo That's in accordance with my question, as I defined irrational exponentation by using limits and continuity. – Poder Rac Aug 03 '20 at 01:38
  • @Angelo Of couse. In this case, the fact that the limits exist implies continuity. – Poder Rac Aug 03 '20 at 02:03
  • I think you have to define irrational exponentation before than you can use the fact that it is a continuous function. First you define irrational exponentation and after you can prove that it is a continuous function. You cannot use continuity of exponentation in order to define irrational exponentation. – Angelo Aug 03 '20 at 02:03
  • @Angelo You define exponentation with limits, show that the limits exist and are unambiguous, which proves continuity. Then you can say that your definition of exponentation hinges on continuity. – Poder Rac Aug 03 '20 at 02:08
  • It is what I said. First definition, then continuity. – Angelo Aug 03 '20 at 02:11
  • @Angelo I see what you mean. Yes, there should have been 'defined by limits' instead of 'defined by continuity' in my post. I will correct that. – Poder Rac Aug 03 '20 at 02:15
  • @PoderRac Out of curiosity: Why do you get yourself in such a weird discussion by being hell-bent on "avoiding derivatives" (in quotes because I don't think that can be properly formalised but ok.) – Maximilian Janisch Aug 03 '20 at 22:41
  • @Maximilian That's a good point. I don't know how would I formalize my requirement at this moment, maybe it could be done somehow (Proof theory?). Nevertheless, mathematicians sometimes use "prove $\text{something}_1$ without using $\text{something}_2$". "Proving the prime number theorem without complex analysis" is an example. – Poder Rac Aug 04 '20 at 21:52
  • @PoderRac I see. I would be surprised if that can be formalised. After all, if you really want to, you can always restate all results that derivatives give you "from first principles" without ever using the word "derivative" . By the way, did you see my answer here? I think it is pretty slick hehe – Maximilian Janisch Aug 05 '20 at 08:18
  • @Maximilian I did. It's a good answer, but still, I prefer Angelo's answer (which I accepted), as it seems to be the most elementary way to prove my inequality. – Poder Rac Aug 05 '20 at 08:26
  • @PoderRac Really? It is completely fine if you choose Angelo's answer but I am not sure if it is more elementary than mine. After all, I only used that the exponential function is convex. – Maximilian Janisch Aug 05 '20 at 08:28
  • @Maximilian In a sense, I would say – it "uses" only the concept of geometric series, which I think is more elementary than proving convexity of functions. – Poder Rac Aug 05 '20 at 08:38
  • @PoderRac I won't disagree because we are really leaving the realm of mathematics here :) – Maximilian Janisch Aug 05 '20 at 09:32
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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.

Angelo
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2

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.

River Li
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1

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 $\;t\in\left]0,1\right[$.

Angelo
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    "Do not use derivatives" – Poder Rac Aug 02 '20 at 21:56
  • You accepted an answer that uses a generalization of Bernoulli’s inequality which can be proved by comparing derivatives. So you accepted an answer that is using derivatives implicitly. So I cannot understand why you accepted that answer but mine do not. – Angelo Aug 02 '20 at 22:15
  • See my comment under cha21's answer. – Poder Rac Aug 02 '20 at 22:16
  • So please show a proof that $\ln x$ is concave without using derivatives and without using AM-GM. I think that in any case derivatives are used implicitly. So I cannot understand why a answer is accepted only because it does not show derivatives explicitly. – Angelo Aug 02 '20 at 22:37
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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 $r=t$.

Maximilian Janisch
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