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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.

Dan
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  • How many roots can there be at most? What value does $x+\log(x)$ take at $x=k+\sqrt{k}$? You might want to plot the variations of the function of interest first. – Anthony Aug 06 '22 at 08:46
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    @AnthonySaint-Criq There are exactly two roots for any $k>1$, one is for $x<1$ and one is for $x>1$. Function is decreasing until $x=1$ then increasing. – Suzu Hirose Aug 06 '22 at 08:57
  • @SuzuHirose I was just suggesting hints towards solving the problem. Of course the function has to be strictly monotonic at some point, and this has to be used! My hint was to look at this root $x_0>1$, and try to compare it to $k+\sqrt{k}$ (or in this case, to $k+\sqrt{k}-1$). – Anthony Aug 06 '22 at 11:34
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    Related: https://math.stackexchange.com/questions/3481810/an-inequality-on-the-roots-of-a-transcendental-equation – River Li Aug 06 '22 at 13:15
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    @RiverLi I would call it "duplicate" rather than "related" – Ewan Delanoy Aug 06 '22 at 13:37
  • @EwanDelanoy There is "(without calculator)". – River Li Aug 06 '22 at 13:40
  • @RiverLi I see. Thanks for your clarification – Ewan Delanoy Aug 06 '22 at 14:08
  • @EwanDelanoy Actually, I don't know if "without calculator" makes difference. :) – River Li Aug 06 '22 at 14:12

1 Answers1

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$f(x)=\log(x)$ 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 $$ W_{-1}(-ue^{-u}) \geq \log u-\sqrt{u-\log u}. $$

Jack D'Aurizio
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  • Thank you. In the first paragraph you wrote $g(k)<0$, but how can $g(x)$ ever be negative? Also, how do we know that the statement in the question is true for $1<k\leq15.4$? – Dan Aug 06 '22 at 22:02
  • Sorry, I meant $g(k)<k$, now fixed. The range $1<k<15.4$ still needs to be settled, I am planning to crack it by proving the last inequality. – Jack D'Aurizio Aug 06 '22 at 22:29