How can I show that $2 < 1/a + 1/b It is given that $b \ln a - a \ln b = a - b$  for $a>0 , b>0 , a \ne b$
I need to show that $$2< 1/a + 1/b < e.$$
I am able to apply laws of logs and therefore introduce $e $ in the expression, but then how do I get $(a+b)/ab$?
 A: Let us write $x:=a/b$. Then the original condition may be written
$$\ln a=\frac{x\ln x}{x-1}-1.$$
We are asked to find the bounds of
$$y:=\frac1a+\frac1b=(1+x)\exp\left(\frac{x\ln x}{1-x}+1\right),$$
where $x>0$ and $x\neq1$. The discontinuity in $y$ at $x=1$ is removable because $\lim_{x\to1-}y=\lim_{x\to1+}y=2$. Also, $\lim_{x\to0+}y=\lim_{x\to\infty}y=\mathrm e$. To find the in-range minimum of $y$, differentiate to get
$$\frac{\mathrm dy}{\mathrm dx}=\frac{2-2x+(1+x)\ln x}{(1-x)^2}\exp\left(\frac{x\ln x}{1-x}+1\right).$$
This is zero if
$$z:=\ln x-2+\frac4{1+x}=0.$$
To see when this happens, consider the gradient
$$\frac{\mathrm dz}{\mathrm dx}=\frac{(1-x)^2}{x(1+x)^2}.$$
This is positive for $x>0$ except at $x=1$, where $z=0$. Hence there is just one minimum for $y$: namely $y=2$ when $x=1$. It follows that $$2<\frac1a+\frac1b<\mathrm e.$$
Edit: $\;$To see that $y\to\mathrm e$ as $x\to\infty$, let $t:=\ln x$. Then
$$\lim_{x\to\infty}y=\lim_{t\to\infty}\left[(1+\mathrm e^t)\exp\left(\frac{t\mathrm e^t}{1-\mathrm e^t}+1\right)\right]\qquad\qquad\qquad\qquad$$
$$=\lim_{t\to\infty}\left[\exp\left(1-\frac t{1-\mathrm e^{-t}}\right)+\exp\left(1-\frac t{\mathrm e^t-1}\right)\right]$$
$$=\mathrm e.$$
A: Hint: Firstly, taking $x=1/a$ , $y=1/b$ so you have $$\frac{y\ln y-x\ln x}{y-x} =1$$
and you have to prove $2< x+y<e$. In order to prove that, you will prove the following inequality:
$$\ln(x+y)<\frac{y\ln y-x\ln x}{y-x} < \ln\left(\frac{x+y}{2} \right)+1$$
A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$

Given that \begin{align}b \ln a - a \ln b = a - b \quad\text{for } a>0, b>0, a\ne b \tag{1}\label{1}\end{align}
show that \begin{align}2< 1/a + 1/b < \e \tag{2}\label{2}\end{align}

Hint.
From \eqref{1} there are two solutions for $b$ in terms of $a$:
\begin{align}
b=f_0(a)&=-\frac{a}{\ln(\e\cdot a)}\,\Wp\left(-\frac{\ln(\e\cdot a)}{\e\cdot a}\right)
\tag{3}\label{3}
,\\
b=f_1(a)&=-\frac{a}{\ln(\e\cdot a)}\,\Wm\left(-\frac{\ln(\e\cdot a)}{\e\cdot a}\right)
\tag{4}\label{4}
,
\end{align}
where $\Wp$ is the principal branch
and $\Wm$ is the other real branch
of the Lambert $\W$ function.

Since the case $a=b$ must be excluded, we need to prove \eqref{2} for two cases:
\begin{align}
1)\quad a\in(\tfrac1\e,1),\quad b=f_0(a)
\tag{5}\label{5}
.
\end{align}
\begin{align}
2)\quad a>1,\quad b=f_1(a)
\tag{6}\label{6}
.
\end{align}
Note that
\begin{align}
\frac11+\frac1{f_0(1)}
&=\frac11+\frac1{f_1(1)}
=2
,\\
\lim_{a\to\tfrac1\e}
\left(\frac1{a}+\frac1{f_1(a)}\right)
&=\e
,\\
\text{and }\quad
\lim_{a\to\infty}
\left(\frac1a+\frac1{f_0(a)}\right)
&=\e
\end{align}


$\endgroup$
