Prove $2 < 1/a + 1/b < \mathrm{e}$ provided $b\ln a - a\ln b = a - b$ 
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.
 A: (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.
A: 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.
A: 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 $x+y<t+e-t=e.$
