Estimate the bound of the sum of the roots of $1/x+\ln x=a$ where $a>1$

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 $$x_1+x_2+1<3\exp(a-1)$$

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 $t=\sqrt{\frac{y}{x}}$, $$ \sqrt{\frac{y}{x}} \ln\frac{y}{x} < \frac{y}{x}-1 = y \bigg(\frac1x-\frac1y\bigg) = y\big((a-\ln x)-(a-\ln y)\big) = y\big(\ln y-\ln x\big) = y \ln\frac{y}{x} \\ xy > 1. $$

I can show the inequality when $a$ is close enough to $1$ (namely, $a\leq 1+ln(5/4)\approx 1.22$) or when $a$ is big enough (namely, $a \geq 1+\ln(5) \approx 2.6$). In the sequel $f(x)$ denotes $x+\ln(\frac{1}{x})$.

When $a$ is close to $1$. Let us put $w=\sqrt{e^{a-1}-1}$. The inequality then becomes $x_1+x_2\leq 2+3w^2$. Now I claim that when $w\in[0,\frac{1}{2}]$, one has

$$ \begin{array}{lcl} x_1 & \leq & 1-\sqrt{2}w+\frac{4}{3}w^2 \\ x_2 & \leq & 1+\sqrt{2}w+\frac{5}{3}w^2 \\ \end{array} $$

To check the first claim, let $g(w)=f(1-\sqrt{2}w+\frac{4}{3}w^2)-1-\ln(1+w^2)$. Then the derivative of $g$ is $g'(w)=\frac{w^3(\frac{4}{3}\sqrt{2}w-\frac{34}{9})}{(1+w^2)(1-\sqrt{2}w+\frac{4}{3}w^2)^2}<0$. So $g$ is decreasing on $(0,\frac{1}{2})$, whence $g(w)\leq g(0)=0$, proving the first claim.

To check the second claim, let $h(w)=f(1+\sqrt{2}w+\frac{5}{3}w^2)-1-\ln(1+w^2)$. Then the derivative of $h$ is $h'(w)=\frac{w^2(\frac{-5}{3}\sqrt{2}w^2-\frac{28}{9}w+sqrt(2)))}{(1+w^2)(1+\sqrt{2}w+\frac{5}{3}w^2)^2}$. So $h'$ has a unique zero in $(0,\frac{1}{2})$ ; since $h(0)=0$ and $h(\frac{1}{2})\approx 0.00091 >0$, we have $h(w)\geq 0$ for any $w\in(0,\frac{1}{2})$, proving the second claim.

When $a$ is big enough. Let us put $u=e^{a-1}$. The inequality then becomes $x_1+x_2+1\leq 3u$. Now I claim that when $u\geq 5$, one has

$$ \begin{array}{lcl} x_1 & \leq & (3-e)u-1 \\ x_2 & \leq & eu \\ \end{array} $$

To check the first claim, let $k(u)=f((3-e)u-1)-1-ln(u)$. Then the derivative of $k$ is $k'(w)=\frac{-1} {u((3-e)u-1)^2}<0$. So $k$ is decreasing on $(0,\infty)$, and since $k(5)\approx -1.05 <0$, this proves the first claim.

The second claim follows from the identity $f(eu)-1-ln(u)=\frac{1}{eu}$.