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Let $f(x)=x-\ln x$. Suppose $f(x_1)=f(x_2)=m$ ($x_1<x_2$). How can I prove that

$$\sqrt{x_1}+\sqrt{x_2}\ge\sqrt{m}+\frac{1}{\sqrt{m}}?$$

My Attempt

I tried to rewrite the condition as \begin{align} &x_1-\ln x_1=m,\\ &x_2-\ln x_2=m. \end{align} By summing up the two equalities and by subtracting one equality from the other, we have \begin{align} &x_1+x_2-\ln x_1x_2=2m,\\ &\sqrt{x_1}+\sqrt{x_2}=\frac{\ln x_2-\ln x_1}{\sqrt{x_2}-\sqrt{x_1}}. \end{align} But I don't know how to continue then.

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xskxzr
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  • in title of question there is $m$? – TheStudent Mar 19 '20 at 04:09
  • @TheStudent It's equivalent. $f(x_1)=f(x_2)=m$ implies $x_1,x_2$ are the roots of $x-\ln x-m$. – xskxzr Mar 19 '20 at 04:13
  • Negative result to report: since we can find $x_1'(m)$ and $x_2'(m)$ by implicit differentiation, I was hoping to prove that the difference between the two sides of the inequality was an increasing function of $m$ ... but computations show that's not actually the case. – Greg Martin Mar 19 '20 at 05:57
  • @GregMartin I think you were on the right track because the LHS - RHS was increasing for $m > \sim 5$ – Varun Vejalla Mar 19 '20 at 06:56
  • It actually starts decreasing again later on :/ , and seems to have limit 0 – Greg Martin Mar 19 '20 at 07:16
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    Some similar problems: https://math.stackexchange.com/questions/3481810/an-inequality-on-the-roots-of-a-transcendental-equation/3483445#3483445, and https://math.stackexchange.com/questions/1725944/estimate-the-bound-of-the-sum-of-the-roots-of-1-x-ln-x-a-where-a1?rq=1 – River Li Mar 19 '20 at 07:23

1 Answers1

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This is true if you consider that $x_1$ and $x_2$ are explicitly given in terms of Lambert function (for $m \geq 1$). $$x_1=-W_0(-e^{-m}) \qquad \text{and} \qquad x_2=-W_{-1}(-e^{-m}) $$

Now, $$x_1^k+x_2^k \geq (m^k+m^{-k})$$ seems to hold for any $k \geq 0.48166\cdots$

Claude Leibovici
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