Let $x_{1},x_{2}$ be two roots of the equation $$x-\ln{x}-m=0\; (m>1).$$ Show that $$1-\ln{m}<x_{1}x_{2}<\dfrac{\ln{m}}{m-1}.$$
So far, I only made $$\begin{cases} x_{1}-\ln{x_{1}}=m\\ x_{2}-\ln{x_{2}}=m \end{cases}\Longrightarrow x_{1}-x_{2}=\ln{\dfrac{x_{1}}{x_{2}}}$$ and set $\dfrac{x_{1}}{x_{2}}=t>1$, then we have $$x_{2}=\dfrac{\ln{t}}{t-1},x_{1}=\dfrac{t\ln{t}}{t-1}.$$ So it remains to prove $$1-\ln{m}<\dfrac{t\ln^2{t}}{(t-1)^2}<\dfrac{\ln{m}}{m-1}.$$
The following discussion follows the approach to express the two inequalities in question as dependent on just one single parameter $s = \frac12 \ln t$, where $t$ has already been introduced by the OP as $\dfrac{x_{1}}{x_{2}}=t>1$. Then one can write the two inequalities as $f(s) > 0$ and $g(s) > 0$ and show these.
Let's call the product $p = x_1 x_2$. As alread posted, $ p =\dfrac{t\ln^2{t}}{(t-1)^2}$, which can be rewritten $ p =\dfrac{s^2}{((\sqrt t-1/\sqrt t )/2)^2} = \dfrac{s^2}{(\sinh(s))^2}$.
Further, consider the general equality $4 x_1 x_2 = (x_1+x_2)^2 - (x_1-x_2)^2$. By the equation in question, $4 x_1 x_2 = (2m + \ln(x_1 x_2))^2 - (\ln(x_1/x_2))^2$ or $$ p = (m + \frac12 \ln(p))^2 - s^2 $$ This gives $m = - \frac12 \ln(p) + \sqrt{p +s^2}$, and, inserting $p$ from above, $m = - \frac12 \ln(\dfrac{s^2}{(\sinh(s))^2}) + \sqrt{\dfrac{s^2}{(\sinh(s))^2} +s^2}$, or simplified,
$$ m = - \ln(s) + \ln(\sinh(s)) + s \coth(s) $$
One observes that as $s=0$, $m(0) = 1$, and $m$ is increasing monotonically and unboundedly with $s$. So we have to consider the positive interval of $s$.
For the left bound we need to establish $f = p - 1 + \ln(m) > 0$. Since $p>0$ anyway, this needs only be considered for $1\leq m < $e . Since at $s= 2.1$, $m(s) >$e, we consider the range $0\leq s<2.1$
Inserting $p$ and $m$ gives
$$ f(s) = \dfrac{s^2}{(\sinh(s))^2} - 1 + \ln(- \ln(s) + \ln(\sinh(s)) + s \coth(s)) $$
This looks as follows:
Further, near $s=0$, by expansion, $f \simeq s^2/6 \geq 0$. This establishes the left inequality.
For the right inequality, we need to establish $g = \frac{ \ln(m) }{m-1} - p> 0$. Inserting $p$ and $m$ gives
$$ g(s) = - \dfrac{s^2}{(\sinh(s))^2} + \frac{\ln(- \ln(s) + \ln(\sinh(s)) + s \coth(s))}{- \ln(s) + \ln(\sinh(s)) + s \coth(s) - 1} $$
Near $s=0$, by expansion, $g \simeq s^2/12 \geq 0$.
The overall behaviour looks as follows:
For larger $s$, we have that $g(s)$ decreases monotonically with the limit $g(s\to \infty) = 0$.
This establishes the right inequality.
$\Box$
For the left inequality.
Let $g(x) := x - \ln x$. $g(x)$ is strictly decreasing on $(0, 1)$, and strictly increasing on $(1, \infty)$. If $m > 1$, $g(x) = m$ has exactly two positive real solutions $0 < x_1 < 1 < x_2$. We need to prove that $x_1 x_2 > 1 - \ln m$.
We only need to prove the case $1 - \ln m > 0$, i.e. $m < \mathrm{e}$.
It is easy to prove that $$g(x) - \frac{x^3 + x^2 + x + 3}{x^2 + 4x + 1}\left\{\begin{array}{cl} < 0 & x > 1 \\ = 0 & x = 1 \\ > 0 & 0 < x < 1. \end{array} \right. \tag{1}$$ Let $$h(x) := \frac{x^3 + x^2 + x + 3}{x^2 + 4x + 1}.$$ It is easy to prove that $h(x)$ is strictly decreasing on $(0, 1)$, and strictly increasing on $(1, \infty)$. Also, we have $h(0) = 3$ and $h(\infty) = \infty$. Thus, if $1 < m < \mathrm{e}$, $h(x) = m$ has exactly two positive real solutions $0 < x_3 < 1 < x_4$. Furthermore, from (1), we have $x_3 \le x_1$ and $x_4 \le x_2$. Thus, we have $x_1 x_2 \ge x_3 x_4$. It suffices to prove that $x_3 x_4 > 1 - \ln m$.
We have $$h(x) = m \iff x^3 - (m-1)x^2 + (1 - 4m)x - (m - 3) = 0.$$ Consider the cubic equation $$F(x) := x^3 - (m-1)x^2 + (1 - 4m)x - (m - 3) = 0.$$ $x_3, x_4$ are two real solutions of $F(x) = 0$. Clearly, the third real solution is negative, denoted by $x_5 < 0$. By Vieta's theorem, we have $x_3 + x_4 + x_5 = m - 1$, $x_3x_4 + x_4x_5 + x_5x_3 = 1-4m$, and $x_3x_4x_5 = m-3$.
Consider the cubic equation $G(u) = 0$ where $$G(u) := (u - x_3x_4)(u - x_4x_5)(u - x_5x_3)$$ $$= u^3 - (1-4m)u^2 + (m-3)(m-1)u - (m-3)^2.$$ Since $x_3x_4 > 0, x_3x_5 < 0, x_4x_5 < 0$. Thus, $G(u) = 0$ has two negative real roots ($x_3x_5, x_4x_5$), and one positive real root $x_3x_4$. We can prove that, for all $1 < m < \mathrm{e}$, $$G(1-\ln m) < 0. \tag{2}$$ (The proof is given at the end.) Thus, we have $x_3x_4 > 1 - \ln m$.
We are done.
$\phantom{2}$
Proof of (2).
We need to prove that, for all $1 < m < \mathrm{e}$, $$-\ln^3 m + (4m+2)\ln^2 m + (-m^2-4m-4)\ln m + 6m-6 < 0.$$
Using $\ln^3 m \ge 2\ln^2 m - \ln m$ (AM-GM), it suffices to prove that $$-(2\ln^2 m - \ln m) + (4m+2)\ln^2 m + (-m^2-4m-4)\ln m + 6m-6 < 0,$$ or $$4m\ln^2 m - (m+1)(m+3)\ln m + 6m - 6 < 0,$$ or \begin{align*} &\frac{m^2 + 4m + 3}{8m} + \frac{3-m}{8m}\sqrt{m^2 + 14m + 1} \\ <{}& \ln m\\ <{}& \frac{m^2 + 4m + 3}{8m} + \frac{3-m}{8m}\sqrt{m^2 + 14m + 1}. \end{align*} Since $\frac{m^2 + 4m + 3}{8m} + \frac{3-m}{8m}\sqrt{m^2 + 14m + 1} > 1 \ge \ln m$, it suffices to prove that $$\ln m > \frac{m^2 + 4m + 3}{8m} + \frac{3-m}{8m}\sqrt{m^2 + 14m + 1} .$$
Using $\frac{m^2 + 4m + 3}{8m} + \frac{3-m}{8m}\sqrt{m^2 + 14m + 1} \le - \frac{3}{25}m^2 + m - \frac{22}{25}$ (easy), it suffices to prove that $$\ln m > - \frac{3}{25}m^2 + m - \frac{22}{25}. \tag{A1}$$ Let $H(m) := \ln m + \frac{3}{25}m^2 - m + \frac{22}{25}$. We have $$H'(m) = \frac{(3m-5)(2m-5)}{25m}.$$ We have $H(1) = 0$, $H(5/3) > 0$, $H(5/2) > 0$, and $H(\mathrm{e}) > 0$. Thus, (A1) is true.
We are done.
For the right inequality. (Sketch of a proof)
Let $g(x) := x - \ln x$. $g(x)$ is strictly decreasing on $(0, 1)$, and strictly increasing on $(1, \infty)$. If $m > 1$, $g(x) = m$ has exactly two positive real solutions $0 < x_1 < 1 < x_2$. We need to prove that $x_1 x_2 < \frac{\ln m}{m - 1}$.
We split into two cases.
Case 1. $m > 5/3$
Since $g(5m/3) - m = \frac23m - \ln(5m/3) > 0$ and $5m/3 > 1$, we have $x_2 < 5m/3$. It suffices to prove that $x_1 < \frac{\ln m}{m-1}\cdot \frac{3}{5m}$. Since $\frac{\ln m}{m-1}\cdot \frac{3}{5m} \in (0, 1)$, it suffices to prove that $$g\left(\frac{\ln m}{m-1}\cdot \frac{3}{5m}\right) - m < 0,$$ or $$\frac{\ln m}{m-1}\cdot \frac{3}{5m} - \ln\left(\frac{\ln m}{m-1}\cdot \frac{3}{5m}\right) - m < 0.$$
Using $\ln m \le \frac{(m-1)(m+1)}{2m}$ and $\ln m \ge \frac{2(m-1)}{m+1}$, it suffices to prove that $$f(m) := \frac{m+1}{2m}\cdot \frac{3}{5m} - \ln\left(\frac{\frac{2(m-1)}{m+1}}{m-1}\cdot \frac{3}{5m}\right) - m < 0.$$ We have $f(5/3) < 0$. It suffices to prove that $f'(m) < 0$ on $m > 5/3$, i.e. $$-\frac{10m^4 - 10m^3 - 7m^2 + 9m + 6}{10m^3(1+m)} < 0$$ which is true.
Case 2. $1 < m \le 5/3$
We can prove that $$x_1 < \frac13 + \frac23 m - \sqrt{2(m-1)}, $$ and $$x_2 < \frac{3(m+1)}{(m^2+4m+1)(\frac13 + \frac23 m - \sqrt{2(m-1)})}.$$
Thus, we have $$x_1 x_2 < \frac{3(m+1)}{m^2 + 4m + 1} < \frac{\ln m}{m-1}$$ where we use $\ln m > \frac{3(m+1)(m-1)}{m^2 + 4m + 1}$ (easy).
We are done.
