TITLE: Estimate the bound of the sum of the roots of $1/x+\ln x=a$ where $a>1$
QUESTION [8 upvotes]: If $a>1$ then  $\frac{1}{x}+\ln x=a$ has two distinct roots($x_1$ and $x_2$, Assume $x_1<x_2$). Show that $$x_1+x_2+1<3\exp(a-1)$$

First I tried to estimate the place of the roots separately. I have got that $x_1\leq \frac{1}{a}$ and $\exp(a-1)<x_2<\exp(a)$. Then I tried to think about $x_1 + x_2$ as a whole. Because $1/x_1+\ln x_1=a$ and $1/x_2+\ln x_2=a$. I tried to express $x_1+x_2$ as a function of $a$, But I failed. I have no idea to solve this problem. Please help me  :)

REPLY [5 votes]: The function $\frac1x+\ln x$ is decreasing in $(0,1]$ and it is increasing in $[1,\infty)$; each has the range $[1,\infty)$. So, we have two inverse functions, $x:[1,\infty)\to (0,1]$ (decreasing) and $y:[1,\infty)\to[1,\infty)$ (increasing), satisfying 
$$ \frac1{x(a)}+\ln x(a)=\frac1{y(a)}+\ln y(a)=a. \tag1$$
We want to prove $x(a)+y(a)+1<3e^{a-1}$ for $a>1$.
As $x(1)=y(1)=1$, we have $x(1)+y(1)+1=3$, so it suffices to prove $\big(\ln(x(a)+y(a)+1)\big)'<1$ for $a>1$.
By differentiating (1), we get $-\frac{x'}{x^2}+\frac{x'}{x}=1$, so
$x'=\frac{x^2}{x-1}$ and similarly $y'=\frac{y^2}{y-1}$, therefore
$$
\big(\ln(x(a)+y(a)+1)\big)'
= \frac{x'+y'}{x+y+1}
= \frac{\frac{x^2}{x-1}+\frac{y^2}{y-1}}{x+y+1} =
1 + \frac{xy-1}{(x-1)(y-1)(x+y+1)}.
$$
In the denominator of the last fraction we have $x<1$, so we need $xy>1$.
By a straightforward calculation it can be verified that $t\ln (t^2)<t^2-1$ for $t>1$. Applying this with $t=\sqrt{\frac{y}{x}}$,
$$
\sqrt{\frac{y}{x}} \ln\frac{y}{x} < \frac{y}{x}-1 
= y \bigg(\frac1x-\frac1y\bigg) 
= y\big((a-\ln x)-(a-\ln y)\big) 
= y\big(\ln y-\ln x\big) 
= y \ln\frac{y}{x} \\
xy > 1.
$$