Proof of $a^x ≥ x+1 \; \forall x \in \Bbb R \implies a=e$

Question

I'm trying to prove the following :

Let $a>0$ a real number. Then : $\quad a^x ≥ x+1 \;\; \forall x \in \Bbb R \iff a=e$

I managed to prove the '$\Longleftarrow$' part : $x≥0$ then $e^x≥x+1$ is clear, knowing that $e^x = \sum\limits_{n≥0} x^n/n!$. On the other hand, if $x<0$, then $1-e^x = \int_x^0 e^tdt ≤ \int_x^0 1 = -x$.

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Here is what I've done for the '$\Longrightarrow$' part. Under the given hypothesis, it is easy to show that $a>1$. Let $f_a(x)=a^x-(x+1) ≥ 0$. Then $$f_a'(x)=\ln(a)a^x-1=0 \iff x=-\log_a(\ln(a))$$ It is easy to show that $m(a) = -\log_a(\ln(a))$ is a minimum for $f_a$. In particular, $$\begin{align} f_a(m(a))≥0 \iff \\ \frac{1}{\ln(a)} - (m(a)+1)≥0 \iff \\ \frac{1}{\ln(a)} ≥ 1- \frac{\ln(\ln(a))}{\ln(a)} \underset{a>1}\iff \\ 1≥\ln(a)-\ln(\ln(a)) \iff \\ e≥\frac{a}{\ln(a)} \iff \\ h(a)=e\ln(a)-a≥0 \end{align}$$

Since $h'(a)=\frac{e}{a}-1=0 \iff a=e$ and $h''(a)<0$, it is easy to see that $h$ has a unique maximum at $a=e$, and then $h(a)≤h(e)=0, h(a)=0 \iff a=e$. Therefore, I conclueded that $a=e$ as desired.

Is it possible to do an easier proof, or is my proof fine ?

Answer 1

Without using derivation, one can use the inequality $\ln(x)\leq x-1$ and its equivalent form that: $$ \ln(x)\geq 1-\frac{1}x. (\star) $$

Now suppose that there is $m$ such that $\forall x$ $\ln(x)\leq m(x-1)$. Note that this is equivalent to your problem. Indeed $m=\ln(a)$ and we can safely assume that $m>0$ (put $x=1$ and then $a>2$). So we have: $$ (1) \ln(x)\leq m(x-1) $$ and using $(\star)$ we have: $$ 1-\frac 1x\leq mx-m \implies mx+\frac 1x\geq m+1. $$ But choosing $x=\frac1{\sqrt m}$ shows that the previous inequality turns into: $$ 2\sqrt m\geq m+1\implies (\sqrt m -1)^2\leq 0. $$ So $m$ should be one.

Answer 2

If you set $t=x+1$, the inequality becomes $a^{t-1}\ge t$ and it's sufficient to analyze it for $t>0$, so it is the same as $$ (t-1)\log a\ge\log t $$

Consider the function $$ f(t)=(t-1)\log a-\log t $$ We want to see whether or not the function is nonnegative. If $0<a\le1$ it's obvious it isn't, since in this case $$ \lim_{t\to\infty}f(t)=-\infty $$ So we assume $a>1$. We have $$ \lim_{x\to0}f(t)=\infty=\lim_{t\to\infty}f(t) $$ and $$ f'(t)=\log a-\frac{1}{t}=\frac{t\log a-1}{t} $$ which shows $f$ has a minimum at $t=1/\log a$. Since $$ f(1/\log a)=1-\log a+\log\log a=\log\frac{e\log a}{a} $$ the function is nonnegative if and only if $$ e\log a\ge a $$

The function $g(u)=e\log u-u$ has derivative $$ g'(u)=\frac{e-u}{u} $$ which vanishes at $u=e$, which is an absolute maximum, and $g(e)=0$. So $e\log a\ge a$ holds if and only if $a=e$.