Inequalities involving zeros of some functions (e.g., $\frac{\ln x}{x}$, $x\ln x$) 
(P1) Let $f(x) = \frac{\ln x}{x}$, $0 < x_1 < x_2$, $a > 0$ and $f(x_1) = f(x_2) = a$. Prove that
$x_2 - x_1 > \mathrm{e}^2\sqrt{\frac{1}{\mathrm{e}a} - 1}$.

The question was posted months ago by @King.Max, and later deleted (voted by some users).
I gave an answer there. According to what @King.Max said, the question
belongs to the high school mathematics (HSM) topic "function and its derivative"
("When I was doing the test, I encountered the problem without any idea."
"Subject name is high school mathematics.")
In this sense, I think that my answer is not nice, and there should be HSM solutions.
See: https://math.stackexchange.com/questions/3593082/prove-an-inequality-about-the-function-ln-x-x
(Note: As @Martin R pointed out, <10K users cannot see deleted Q&As. They may not see this link. I gave my answer in the link as an answer for this post.)
I have seen many similar problems. These problems has the following description:
Let $f(z)$ be a unimodal function. Let $x \ne y$ with $f(x) = f(y) = a$. Then $g(x, y) \ge h(a)$ for two functions $g(x,y)$ and $h(a)$.
There may be some common methods for the class of problems. See my answer to this question Prove $(x-1)(y-1)>(e-1)^2$ where $x^y=y^x$, $y>x>0$.
Any comments and solutions are welcome and appreciated.
I give some similar problems here (Problems 2-9).

 >(P2) Let $f(x)=x\ln x$, $0 < x_1 < x_2 < 1$, and $f(x_1) = f(x_2) = m$. Prove that: i) $x_2 - x_1 > \sqrt{m\mathrm{e} + 1}$; ii) $\frac{x_1}{x_2} < -m\mathrm{e}$; iii) $\frac{x_1}{x_2} < \frac{-m\mathrm{e}}{2+m\mathrm{e}}$;iv) $\frac{x_1}{x_2} < \frac{-m\mathrm{e}}{10+9m\mathrm{e}}$.


 >(P3) Let $f(x) = (x-1)\ln x$, $0 < x_1 < x_2$, and $f(x_1) = f(x_2) = m$. Prove that $\frac{9}{5} + \ln(1+m) \le x_1 + x_2 \le 2 + \frac{m}{2}$. See: inequality related to roots of $(x-1)\log(x)=m$


 >(P4) Let $f(x) = (x-1)\ln x$, $0 < x_1 < x_2$, $f(x_1) = f(x_2)$. Prove that $\frac{1}{\ln x_1}+ \frac{1}{\ln x_2} < \frac{1}{2}$. See: let $f(x) = (x-1)\ln x$, and given $0 < a < b$. If $f(a) = f(b)$, prove that $\frac{1}{\ln a}+\frac{1}{\ln b} < \frac{1}{2}$


 >(P5) Let $f(x) = x - \ln x$, $0 < x_1 < x_2$, $a > 1$, and $f(x_1) = f(x_2) = a$. Prove that $x_1 + x_2 \le a + \sqrt{a}$.   See: An inequality on the roots of a transcendental equation


 >(P6) Let $f(x) = x - \ln x$, $0 < x_1 < x_2$, and $f(x_1) = f(x_2) = m$. Prove that $\sqrt{x_1} + \sqrt{x_2} \ge \sqrt{m} + \frac{1}{\sqrt{m}}$.  See: Lower bound for the square root sum of the roots of $x - \ln x - m$


 >(P7) Let $f(x) = \frac{1}{x} + \ln x$, $0 < x_1 < x_2$, $a > 1$, and $f(x_1) = f(x_2) = a$. Prove that $x_1 + x_2 + 1 < 3\mathrm{e}^{a-1}$.  See: Estimate the bound of the sum of the roots of $1/x+\ln x=a$ where $a>1$


 >(P8) Let $f(x) = \sin x + \sin \frac{x}{2}$, $0 < x_1 < x_2 < 2\pi$, $m > 0$, and $f(x_1) = f(x_2) = m$. Prove that $\frac{1}{x_1} + \frac{1}{x_2} > \frac{3}{2m}$.  (P8 and P9 were posted by @GiangNguyễnĐặngThanh, the user was removed)


 >(P9) Let $f(x) = \sin x + \sin \frac{x}{2} + \sin \frac{x}{3}$, $m > 0$ and $0 < x_1 < x_2 < 3\pi$, and $f(x_1) = f(x_2) = m$. Prove that $\frac{\pi}{x_1} + \frac{\pi}{x_2} < \frac{8}{m}$.

 A: The copy of my answer for P1 in the link below. (As @Martin R pointed out, < 10K users may not see it.)
See: https://math.stackexchange.com/questions/3593082/prove-an-inequality-about-the-function-ln-x-x
Remark: Since it is a problem of high school mathematics (HSM) topic "function and its derivative", the solution is supposed to be basic and simple. The following solution is not simple.
We have $f'(x) = \frac{1 - \ln x}{x^2}$. Since $f'(\mathrm{e}) = 0$ and $g(x) = \ln x$ is strictly increasing,
we know that $f(x)$ is strictly increasing on $(0, \mathrm{e})$, and strictly decreasing on $(\mathrm{e}, \infty)$.
Also, $f(1) = 0$, $f(x) < 0$ for $x < 1$, and $f(x) > 0$ for $x > 1$.
Thus, $0 < a < \frac{1}{\mathrm{e}}$ and $1 < x_1 < \mathrm{e} < x_2$.
Moreover, if $f(b) > a$, then $x_1 < b < x_2$.
From $\ln x_1 = a x_1$ and $\ln x_2 = ax_2$, we have $a(x_2 - x_1) = \ln \frac{x_2}{x_1}$, and
$\mathrm{e}^{a(x_2-x_1)} = \frac{x_2}{x_1}$, and $\mathrm{e}^{a(x_2-x_1)} - 1 = \frac{x_2 - x_1}{x_1}$,
and
$$a x_1 = \frac{a(x_2 - x_1)}{\mathrm{e}^{a(x_2-x_1)} - 1}.\tag{1}$$
Fact 1: $h(u) = \frac{u}{\mathrm{e}^{u} - 1}$ is strictly decreasing on $(0, \infty)$.
Denote $C = \mathrm{e}^2\sqrt{\frac{1}{\mathrm{e}a} - 1}$. From Fact 1 and (1), we have
\begin{align}
a x_1 < h(aC) \quad
\Longrightarrow \quad  x_2 - x_1 > C.
\end{align}
Thus, it suffices to prove that
$$x_1 < \frac{h(aC)}{a}.$$
Recall that if $f(b) > a$, then $x_1 < b < x_2$. It suffices to prove that
$$f(\tfrac{h(aC)}{a}) > a$$
or
$$\ln h(aC) - h(aC) > \ln a.\tag{2}$$
Fact 2: $\ln h(u) - h(u) \ge -\frac{1}{8}u^2 - 1$ for $u > 0$.
From Fact 2 and (2), it suffices to prove that
$$-\frac{1}{8}(aC)^2 - 1 > \ln a.$$
Since $\mathrm{e}^2 < 8$, it suffices to prove that
$$-\frac{1}{\mathrm{e}^2}(aC)^2 - 1 > \ln a$$
or
$$(a\mathrm{e})^2 - (a\mathrm{e}) - \ln (a\mathrm{e}) > 0.$$
Since $0 < a\mathrm{e} < 1$, it suffices to prove that $x^2 - x - \ln x > 0$ for $0 < x < 1$. Easy. We are done.
$\phantom{2}$
Proof of Fact 1: We have
$h'(u) = -\frac{1}{(\mathrm{e}^u - 1)^2}(\mathrm{e}^uu - \mathrm{e}^u + 1)$.
Let $h_1(u) = \mathrm{e}^uu - \mathrm{e}^u + 1$.
We have $h_1'(u) = \mathrm{e}^uu > 0$ for $u > 0$.
Also, $h_1(0) = 0$. Thus, $h_1(u) > 0$ for $u > 0$.
Thus, $h'(u) < 0$ for $u > 0$. Thus, $h(u)$ is strictly decreasing on $(0, \infty)$. We are done.
Proof of Fact 2: Let $F(u) = \ln h(u) - h(u) + \frac{1}{8}u^2 + 1$. We have
$F'(u) = \frac{(u\mathrm{e}^u - 2\mathrm{e}^u + u+2)^2}{4u(\mathrm{e}^u - 1)^2} \ge 0$ for $u > 0$.
Also, $F(0+) = 0$ (since $h(0+) = 1$). We are done.
