How can I show that $2 < 1/a + 1/b

Question

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 $(a+b)/ab$?

Answer 1

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 $$\lim_{x\to\infty}y=\lim_{t\to\infty}\left[(1+\mathrm e^t)\exp\left(\frac{t\mathrm e^t}{1-\mathrm e^t}+1\right)\right]\qquad\qquad\qquad\qquad$$

$$=\lim_{t\to\infty}\left[\exp\left(1-\frac t{1-\mathrm e^{-t}}\right)+\exp\left(1-\frac t{\mathrm e^t-1}\right)\right]$$ $$=\mathrm e.$$

Answer 2

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: $$\ln(x+y)<\frac{y\ln y-x\ln x}{y-x} < \ln\left(\frac{x+y}{2} \right)+1$$

Answer 3

$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$

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}

$\endgroup$