Prove or disprove: { sum of real roots of $x-\ln{x}=k$ } $1$. (without calculator) Prove or disprove: { sum of real roots of $x-\ln{x}=k$ } $<k+\sqrt{k}$ for $k>1$. (without calculator)
It seems true based on inspection.
My attempt:
Let the real roots of $x-\ln{x}=k$ be $a$ and $b$, $a<b$.
I tried to find functions $f(x)$ and $g(x)$ such that:

The root of $f(x)=k$, call it $\alpha$, is obtainable and slightly greater than $a$.
The root of $g(x)=k$, call it $\beta$, is obtainable and slightly greater than $b$.
$\alpha+\beta<k+\sqrt{k}$

But I haven't been able to find such functions.
The question comes from my former high school student.
 A: $f(x)=\log(x)$ is a concave function on $\mathbb{R}^+$ and $y=x-1$ is the equation of the tangent line at $(1;0)$.
It follows that for any $k>1$ the equation $f(x)=x-k$ has two solutions, one in the interval $(0,1)$ and the other one in the interval $(1,+\infty)$. Let us focus on the largest one $\zeta_k>1$ and consider $g(x)=x-\log(x)$. $g(x)$ is a convex, increasing function and $g(k)<k$, implying that $\zeta_k > k$. By one step of Newton's method we get
$$ \zeta_k < k + \frac{\log k}{1-\frac{1}{k}} $$
so the sum of the roots is controlled by $1+k+\frac{k\log k}{k-1}$. This bound is tighter than $k+\sqrt{k}$ for any $k\geq 15.4$.
In terms of the Lambert function the sum of the roots is explicitly given by
$$ -W_0(-e^{-k})-W_{-1}(-e^{-k}) $$
so the claim is equivalent to the following statement: for any $z\in(0,e^{-1})$
$$ -W_0(-z)-W_{-1}(-z)\leq -\log z+\sqrt{-\log z} $$
or: for any $s\in(-e^{-1},0)$
$$ -W_0(s)-W_{-1}(s) \leq -\log(-s)+\sqrt{-\log(-s)}. $$
Let $s= te^{t}$ for $t\in(-1,0)$. The claim becomes that over such interval
$$ -W_{-1}(te^t) \leq -\log(-t)+\sqrt{-t-\log(-t)}$$
or that for any $u\in(0,1)$ we have
$$ W_{-1}(-ue^{-u}) \geq \log u-\sqrt{u-\log u}. $$
