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Let $f(x)=\dfrac{x}{e^x}$, if $f(a)=f(b)$, $a \neq b$, find the maximum of $\lambda$ such that $a+b+\lambda \ln ab >2$ holds always.

My Attempt.

A method to eliminate the variable is $a=tb$, substituting this into $f(a)=f(b)$ gives $a=\dfrac{\ln t}{t-1}, b=\dfrac{t\ln t}{t-1}$, substituting these into the inequality: $$\dfrac{(t+1)\ln t}{t-1}+\lambda \ln \dfrac{t\ln^2 t}{(t-1)^2}>2$$

It's HARD to calculate out the maximum value of $\lambda$.

An inequality that has been proven is $a+b+\ln ab>2$, which can be split into $a+\ln b>1, b+\ln a>1$ to prove. We can construct $F(x)=f(x)-f(1-\ln x)=\dfrac{x^2-(1-\ln x)e^{x-1}}{xe^x}$, where the numerator is negative in $(0,1)$ and positive in $(1,+\infty)$, and use $F(a)<0, F(b)>0$ and the monotonicity of $f(x)$ to prove .$(0<a<1<b)$

Thinh Dinh
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SlashQwerty
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2 Answers2

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Let $a(t)=\frac{\ln t}{t-1}$ and $b(t)=\frac{t\ln t}{t-1}$. We want to find the largest $\lambda$ which strictly exceeds $$g(t):=\frac{a(t)+b(t)-2}{-\ln(a(t)b(t))}$$ for $t>1$. We claim that $g(t)$ is decreasing in $t>1$. To show this, it suffices to show that $a(t)+b(t)$ is increasing in $t$, while $a(t)b(t)$ is decreasing. Indeed, \begin{align*} \frac{d}{dt}\big(a(t)+b(t)\big)&=\frac{t^2-2t\ln t-1}{t(t-1)^2}\\ \frac{d}{dt}\big(a(t)b(t)\big)&=-\frac{\ln(t)\big((t+1)\ln t-2(t-1)\big)}{(t-1)^3}, \end{align*} so it suffices to show that $t^2-2t\ln t-1>0$ and $(t+1)\ln t-2(t-1)>0$ for all $t>1$. Both of these attain equality at $t=1$, so it's enough to show that the left sides of each are increasing in $t$. That is, we need \begin{align*} 0<\frac d{dt}(t^2-2t\ln t-1)=2t-2\ln t-2=2(t-1-\ln t)\\ 0<\frac d{dt}((t+1)\ln t-2(t-1))=\frac1t+\ln t-1. \end{align*} The positivity of the first is the classic inequality $\ln t<t-1$. For the second, we note that $f(t)=\frac1t+\ln t-1$ is zero at $t=1$, and so it's enough to show that $f'(t)>0$. Indeed $f'(t)=\frac1t-\frac1{t^2}>0$.

Now, $g(t)$ is decreasing in $t>1$, so our answer will be $$\lambda=\lim_{t\to 1^+}\frac{a(t)+b(t)-2}{-\ln(a(t)b(t))}.$$ For $t=1+\varepsilon$, we have $\ln t = \varepsilon-\varepsilon^2/2+\varepsilon^3/3+O(\varepsilon)$, and so \begin{align*} a(t)&=1-\frac{\varepsilon}2+\frac{\varepsilon^2}3+O(\varepsilon^3)\\ b(t)&=1+\frac{\varepsilon}2-\frac{\varepsilon^2}6+O(\varepsilon^3). \end{align*} So $a(t)+b(t)-2=\varepsilon^2/6+O(\varepsilon^3)$ and $a(t)b(t)=1-\varepsilon^2/12+O(\varepsilon^3)$, from which we conclude $\ln(a(t)b(t))=-\varepsilon^2/12+O(\varepsilon^3)$. This gives $\lambda=2$.

Carl Schildkraut
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  • I wonder what we could do using Lambert function. Any idea ? – Claude Leibovici Apr 20 '25 at 01:44
  • @ClaudeLeibovici I'd doubt that the Lambert function would do anything useful here, since the solutions to $a/e^a=b/e^b$ can be directly parametrized in terms of simple functions. Do you have any intuition to the contrary? – Carl Schildkraut Apr 20 '25 at 02:55
  • To the contrary, absolutely not. My idea was that if $\dfrac{a}{e^a}=\dfrac{b}{e^b}=k$, then $a=-W_0(-k)$ and $b=-W_{-1}(-k)$ or vice versa and then use the parametrization of Lambert function as in https://math.stackexchange.com/questions/3233107/parametric-representation-of-the-real-branches-operatornamew-0-operatorna – Claude Leibovici Apr 20 '25 at 03:19
  • The transformation you made in the original answer to the original inequality is actually $\dfrac{a(t)+b(t)-2}{\ln[a(t)b(t)]} < -\lambda$, I think we should discuss the limit at $0$ or $+\infty$,but your approach is very effective. – SlashQwerty Apr 20 '25 at 10:52
  • @SlashQwerty Ah, thanks for pointing that out. I've fixed the sign errors. – Carl Schildkraut Apr 21 '25 at 08:34
  • @CarlSchildkraut You should check that if $\frac{(t+1)\ln t}{t-1}+2 \ln \frac{t\ln^2 t}{(t-1)^2}>2$ is true. – River Li Apr 21 '25 at 23:25
  • @RiverLi Don't I do this? Since $\frac{t\ln^2t}{(t-1)^2}<1$ (this follows from the fact that $a(t)b(t)$ is decreasing in $t$), this is equivalent to $$\frac{\frac{(t+1)\ln t}{t-1}-2}{-\ln\frac{t\ln^2t}{(t-1)^2}}>2.$$ The left side is decreasing in $t$ (first part of my answer), and the limit of it as $t\to 1$ is $2$. So the inequality in your comment never holds. – Carl Schildkraut Apr 21 '25 at 23:31
  • @CarlSchildkraut If $t=2$, $\frac{(t+1)\ln t}{t-1}+2 \ln \frac{t\ln^2 t}{(t-1)^2}-2 < 0$. Can you check it? – River Li Apr 21 '25 at 23:37
  • @CarlSchildkraut What is your opinion on my last comment? – River Li Apr 28 '25 at 02:40
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Some thoughts.

The OP has obtained $$\dfrac{(t+1)\ln t}{t-1}+\lambda \ln \dfrac{t\ln^2 t}{(t-1)^2}>2,$$ or $$\dfrac{(t+1)}{t-1}+\frac{\lambda}{\ln t}\cdot \ln \dfrac{t\ln^2 t}{(t-1)^2}>\frac{2}{\ln t}.\tag{1}$$ Taking limit $t \to \infty$ on (1), we have $1 - \lambda \ge 0$. Thus, $\lambda \le 1$.

River Li
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