3

Problem: Let $f(x) = x (1 - \ln x)$.
(1) Discuss the monotonicity of $f$.
(2) Let $a, b > 0$ with $a \ne b$ and $b\ln a - a\ln b = a - b$. Prove that $2 < 1/a + 1/b < \mathrm{e}$.

Source: China's national college entrance exam (a.k.a. Gaokao), math exam, June 7, 2021. (It was reported that in 2021 more than 10 million students took Gaokao.)

Remark 1: It was posted partly in (now CLOSED): How can I show that $2 < 1/a + 1/b <e$?

Remark 2: Since Gaokao assesses high school students, elementary solutions are preferred (e.g. without using the Lambert W function).

Remark 3: There are many similar problems in MSE or AoPS. These problems have the following description:
Let $f(x)$ be a unimodal function. Let $f(a) = f(b) = m$ for some $0 < a < b$. Prove that $g(a, b) \ge h(m)$.

Indeed, in this problem, $b \ln a - a \ln b = a - b$ is written as $$\frac{1 + \ln a}{a} = \frac{1 + \ln b}{b}.$$ Just let $f(x) = \frac{1 + \ln x}{x}$, $g(a, b) = \frac{1}{a} + \frac{1}{b}$ and $h(m) = 2$ etc.

See: let $f(x) = (x-1)\ln x$, and given $0 < a < b$. If $f(a) = f(b)$, prove that $\frac{1}{\ln a}+\frac{1}{\ln b} < \frac{1}{2}$,
and Prove $(x-1)(y-1)>(e-1)^2$ where $x^y=y^x$, $y>x>0$.,
and Inequalities involving zeros of some functions (e.g., $\frac{\ln x}{x}$, $x\ln x$) (CLOSED).

Hope to see nice proofs (even elegant proofs).

Any comments and solutions are welcome and appreciated.

River Li
  • 37,323
  • 1
    It's good that you give source of your question. However you didn't mention which region it was - as different regions have different versions of the exam. – WhatsUp Jun 10 '21 at 01:32
  • 1
    @WhatsUp Thanks. It is New Gaokao volume I. – River Li Jun 10 '21 at 01:56

3 Answers3

2

(1) is easy: $f'(x)=-\ln(x)$. Thus $f(x)$ is increasing on $(0,1)$ and decreasing on $(1,+\infty)$. Moreover, $f(x)>0$ on $(0,1)$ and $(1,e)$.

(2) The given condition says that $f(1/a)=f(1/b)$. For simplicity, write $y=1/a,z=1/b$. WLOG, assume $y<z$. Thus $0<y<1<z<e$. The inequality $y+z>2$ can be rewritten to $1>y>2-z$. It suffices to consider when $2-z>0$. By (1), $y>2-z$ is equivalent to $f(z)=f(y)>f(2-z)$. Consider $g(z)=f(z)-f(2-z)$. It suffices to show that $g(z)>0$ when $z\in (1,2)$. This is easy since $g'(z)=-\ln(1-(z-1)^2)$. This shows $y+z>2$.

The other half is similar. Here are some details. Note $y+z<e$ is equivalent to $y<e-z$. If $e-z\ge 1$, this obvious. Now assume that $e-z< 1$, which implies that $z\in (e-1,e)$. Then we need to show that $0<y<e-z<1$, or $f(y)=f(z)<f(e-z)$ for $z\in (e-1,e)$. Consider $h(z)=f(e-z)-f(z)$, $z\in (e-1,e)$. We have $h'(z)=\ln(z(e-z))$, which has a critical point at $z_0=\frac{e+\sqrt{e^2-4}}{2}$.One can check that $h(z)$ is increasing on $(e-1,z_0)$ and decreasing on $(z_0,e)$. Thus it suffices to show that $h(e-1)\ge 0$ and $h(e)\ge 0$. But that is clear.

Q-Zhang
  • 1,608
  • 1
    It is nice. You may not put everything in one paragraph. – River Li Jun 10 '21 at 02:12
  • @SeeHai See the new edit. – Q-Zhang Jun 10 '21 at 14:57
  • Thank you! But I have 2 questions: Since calculators are not allowed on the Gaokao, how would you show that $h(e-1) \geq 0$? Secondly, I don't think $h(e)$ is defined since $f$ is not defined at $0$? – Hello Jun 10 '21 at 15:50
  • 1
    (1) Isn't it obvious? $h(e-1)=f(1)-f(e-1)$, but $f$ is decreasing on $(1,e)$. （2）Of course $f(0)$ is interpreted as $\lim_{x\to 0^+}f(x)$. – Q-Zhang Jun 10 '21 at 16:05
  • Ah right, that is obvious for $h(e-1), $ silly me. But you would still need L'Hopital's Rule for $f(0)$? Unless there is a more elementary method I'm not aware of? – Hello Jun 10 '21 at 16:14
  • 1
    @SeeHai One can use L'H. But that is very standard. By the way, a more conceptual proof is added. – Q-Zhang Jun 10 '21 at 16:50
  • Thank you for the clarifications! – Hello Jun 10 '21 at 16:59
  • 1
    with your $ 0<y<c<1<d<z<e $ and, at this point, I guess you are already requiring $f(y) = f(z),$ I do not see why $f'(c) + f'(d) >0 .$ It is true that $f(c) + f(d) > 0$ as both are positive. For example, replace everything with (new letters) $ \psi(x) = 1 - x^2 $ and $-1 < y < c < 0 < d < z < 1.$ If we demand $ \psi(y) = \psi(z) $ we simply have $y+z=0;$ for this example problem we have exact $ \psi'(y) + \psi'(z) = 0 $ Small perturbations of $\psi,$ say $\psi_\delta(x) = \psi (x) + \delta \omega(x),$ may cause $ \psi'(y) + \psi'(z) $ to be positive or negative ... – Will Jagy Jun 10 '21 at 17:58
  • @WillJagy You are right. – Q-Zhang Jun 11 '21 at 00:59
  • we can take $g(x) = f(x) - f(e-x)$ and note that, for $0 < x < e/2,$ we find $g''(x) < 0,$ that is "concave down." Also $g(0^+) = g(e/2) = 0.$ – Will Jagy Jun 11 '21 at 01:33
2

The part $2 < \dfrac{1}{a}+ \dfrac{1}{b}$ has already been proven. We give $2$ different proofs of the other part. The first proof builds upon the idea given by @Q. Zhang, but uses more advanced calculus tools.

Let $x=\dfrac{1}{a} \in (0,1)$ and $y=\dfrac{1}{b} \in (1,e)$. We have $f(x)=f(y)$. Now, \begin{align} x+y <e & \iff y<e-x \\ & \iff f(x)=f(y) >f(e-x). \end{align}

Let $g(x)=f(x)-f(e-x)$, so $g'(x)=-\ln(x)-\ln(e-x)=-\ln[x(e-x)]$. We have to prove that $g(x)>0 \ \forall \ x \in (0,1)$. We have:

\begin{align} g'(x) > 0 & \iff x(e-x)<1 \\ & \iff x<\dfrac{e-\sqrt{e^2-4}}{2}<1 \ \text{or} \ x > \dfrac{e+\sqrt{e^2-4}}{2}>1. \end{align}

Hence, $g'(x) \geq 0$ on $\left(0,\dfrac{e-\sqrt{e^2-4}}{2}\right]$, i.e. $g(x)$ is increasing on this interval. It hence suffices to prove $\lim_{x \rightarrow 0^+} g(x)=0.$ But it is easy to see that $\lim_{x \rightarrow 0^+} g(x)= \lim_{x \rightarrow 0^+} (-x\ln x) $, which is (famously) equal to zero by L'Hopital's Rule.

On the other hand, $g'(x)<0$ on $\left(\dfrac{e-\sqrt{e^2-4}}{2},1\right)$, i.e. $g(x)$ is strictly decreasing on this interval. In this case, it then suffices to show that $g(1)=1-(e-1)[1-\ln(e-1)]>0$. To this end, consider the function $h(x)=1-x(1-\ln x)$, with derivative $h'(x)=\ln x$. Then it is clear that $h(x)$ is strictly increasing for $x\geq1$, and we have $h(e-1)=g(1)>h(1)=0$, and we are done.

Hello
  • 2,133
  • enough to point out that $g''(x) < 0$ for $0<x< \frac{e}{2} ,$ called "concave down" in some books – Will Jagy Jun 10 '21 at 18:23
  • http://macs.citadel.edu/chenm/131.dir/03sp.dir/lect3_5.pdf – Will Jagy Jun 10 '21 at 18:52
  • Nice. The problem boils down to proving $f(x)-f(e-x)\ge 0$ for all $0 < x < 1$. Your solution is quite standard. Perhaps there is no more elementary approach. – River Li Jun 11 '21 at 03:59
2

Second solution. This is more of a geometric approach. Let $f(x)=f(y)=t, 0<t<1.$

Claim $1$: $x<t.$

Proof: Construct the line segment connecting $(0,0)$ to $(1,1)$, with equation $g(x)=x,0<x<1.$ We have $f(x)>g(x) \ \forall \ x \ \in (0,1).$ This gives us $g(t)=t=f(x)>g(x) \Rightarrow t>x,$ since $g$ is monotonic increasing.

Claim $2$: $y<e-t.$

Proof: Construct the tangent to $f(x)$ at $x=e.$ It is easy to see that this tangent has an equation $h(x)=-x+e$. Define $l(x)=h(x)-f(x)$, with derivative $l'(x)=\ln x-1$. Clearly, $l(x)$ is decreasing on $(1,e)$, and since $l(e)=0, h(x)>f(x) \ \forall \ x \in (1,e).$ Now, $(e-t,t)$ is a point on $h(x)$, so we have $h(e-t)=t=f(y)<h(y) \Rightarrow e-t >y,$ since $l$ is monotonic decreasing.

Hence, summing up the $2$ inequalities gives us $x+y<t+e-t=e.$

Hello
  • 2,133