Bound on $x$ for $x-\ln x\ge 1+\epsilon$

Question

I'm looking for a function $x(\cdot)$ where the domain is $\mathbb{R}^+$, that satisfies the following:

1. For all $\epsilon > 0$, the inequality $x-\ln x\ge 1+\epsilon$ is satisfied for all $x \geq x(\epsilon)$, and

2. For all $\epsilon \in \mathbb{R}^+$, the value of $x(\epsilon)$ is close to being as small as possible while still satisfying 1. above.

For example, for all $\epsilon > 0$, the inequality $x-\ln x\ge 1+\epsilon$ clearly is satisfied for all $x \geq 100+100\epsilon$. But given any such $\epsilon$, as $x$ can still satisfy the inequality $x-\ln x\ge 1+\epsilon$ and be much smaller than $100+100\epsilon$, the function $100+100\epsilon$ is not what would suffice here for $x(\epsilon)$.

Furthermore, for such an $x(\epsilon)$ I need a close-form expression.

I've tried playing with iteratively setting $x\ge\ln x + 1+\epsilon$, but I don't see it leading anywhere: $x\ge\ln x+ 1+\epsilon\ge\ln (\ln x+ 1+\epsilon)+ 1+\epsilon\ge\ln (\ldots (\ln x+ 1+\epsilon)\ldots+ 1+\epsilon)+ 1+\epsilon$.

Intuitively, it should suffice that $x$ will be slightly larger than $1+\epsilon$, but I'm struggling to get a reasonable value for it.

Any ideas?

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EDIT: Playing a bit with Wolfram Alpha, I saw that setting $x(\epsilon)=1.6(1+\epsilon)$ also works. This is not ideal as I hope for a bound that tends to $1$ as $\epsilon\to 0$.

EDIT2: It seems that $x(\epsilon)=1+\epsilon+\sqrt {2\epsilon}$ also works. I don't have any intuition as to why this works or if there's an easy way of proving it.

In contrast, it seems that $1+c\cdot \epsilon$ doesn't work for any $c\in\mathbb R$.

Answer 1

$x(\epsilon)=x-\ln(x)-1$ is the largest upper bound possible, and also happens to be in closed form.

Answer 2

First, we observe that $$ f(x)=x-\ln x -1\geq \frac{(x-1)^2}{2}-\frac{(x-1)^3}{3}=g(x). $$ Now clearly if $$ g(x)\geq\varepsilon \Rightarrow f(x)\geq \varepsilon $$

Now all we have to do is to find the roots $f(x)-\varepsilon=0$ and identify the one which is larger than, but close to $1$. It is given by $$ x(\varepsilon)=-\frac{1}{4}(1-12\varepsilon+2\sqrt{36\varepsilon^2-6\varepsilon})^\tfrac{1}{3}-\frac{1}{4(1-12\varepsilon+2\sqrt{36\varepsilon^2-6\varepsilon})^\tfrac{1}{3}}+\frac{3}{2}-\frac{i\sqrt{3}}{2}\left(\frac{1-12\varepsilon+2\sqrt{36\varepsilon^2-6\varepsilon})^\tfrac{1}{3}}{2}-\frac{1}{2(1-12\varepsilon+2\sqrt{36\varepsilon^2-6\varepsilon})^\tfrac{1}{3})}\right) $$

edit: If this expression does not satisfy you, you can of, of course expand it around $\varepsilon=0$ to get:

$$ x(\varepsilon)\approx 1+\sqrt{2\varepsilon}+\tfrac{2}{3}\varepsilon+\tfrac{5}{9}\sqrt{2}\varepsilon^\tfrac{3}{2} $$