This is a question that I came across a few days ago，Although it is not particularly like a research problem, the following problem is that I study the zero distribution of a class of elementary transcendental functions.And I don't think the following problems are easy to deal with. I 've been thinking about them for a few days and they 've all failed



> 
> if $f(x)=x^2-x-\ln{x}-\ln{a}$,and $f(x\_{1})=f(x\_{2})=0,0<x\_{1}<x\_{2}$.I have conjecture the roots $x\_{1},x\_{2}$ such
>  $$\dfrac{3}{2a+1}<x\_{1}x\_{2}<\dfrac{\ln{a}}{a-1}，a>1$$
> 
> 
> 


**This is my attempt**


since
$$x^2\_{1}-x\_{1}-\ln{x\_{1}}=x^2\_{2}-x\_{2}-\ln{x\_{2}}=\ln{a}$$
let $x\_{2}=tx\_{1},t>1$,then we have
$$x^2\_{1}-x\_{1}-\ln{x\_{1}}=t^2x^2\_{1}-tx\_{1}-\ln{t}-\ln{x\_{1}}$$
$$x\_{1}=\dfrac{t-1+\sqrt{(t-1)^2+4\ln{t}\cdot(t^2-1)}}{2(t^2-1)}=f(t)$$
then
$$x\_{2}x\_{1}=t(x\_{1})^2=t\left(\dfrac{t-1+\sqrt{(t-1)^2+4\ln{t}\cdot(t^2-1)}}{2(t^2-1)}\right)^2=t(f(t))^2$$
and 
$$a=e^{x^2\_{1}-x\_{1}-\ln{x\_{1}}}=e^{f^2(t)-f(t)-\ln{f(t)}}$$
so it must prove
$$\dfrac{3}{2e^{f^2(t)-f(t)-\ln{f(t)}}+1}\le t(f(t))^2<\dfrac{f^2(t)-f(t)-\ln{f(t)}}{e^{f^2(t)-f(t)-\ln{f(t)}}-1},t>1$$Next thing I know, it's pretty complicated.



Logarithmic and polynomial functions with two roots
$\newcommand{\ep}{\varepsilon}$
Each of the two proposed inequalities can be written in the form $H(a)>0$ for $a\in(1,\infty)$, where $H$ is a smooth function. All such inequalities -- but only for $a$ in a finite closed interval $I$ -- can be proved, at least in principle, by the interval method; that is, by partitioning $I$ into many small subintervals and bounding, tightly enough, terms in the expression of $H(a)$ on each small subinterval. A difficulty of using the interval method here is that $x\_1$ and $x\_2$ as functions of $a$ are known only implicitly. Another difficulty is that the interval $I$ is neither finite nor closed, and so, we shall need to consider separately ("small") values of $a$ close to $1+$ and large values of $a$, close enough to $\infty$. For intermediate values of $a$, we shall indeed use the interval method. 


Therefore, the solution presented below is long and technical. Since the problem does not seem to belong to a general class of problems, I'd be surprised if there is an "official" solution to it. 


**All and, in particular, large values of $a$.** Let 
\begin{equation\*}
 g(x):=x^2-x-\ln x
\end{equation\*}
for $x>0$. Then $g(1)=g'(1)=0$ and $g$ is strictly convex, with $g(0+)=g(\infty-)=\infty$. So, $g$ is strictly decreasing from $\infty$ to $0$ on $(0,1]$ and strictly increasing from $0$ to $\infty$ on $[1,\infty)$. So, for each real $a>0$ there are uniquely determined $x\_1=x\_1(a)\in(0,1)$ and $x\_2=x\_2(a)\in(1,\infty)$ such that 
\begin{equation\*}
 g(x\_1)=g(x\_2)=\ln a; 
\end{equation\*}
of course, these $x\_1$ and $x\_2$ are the same as in the OP. 


One may note that, if $a\to\infty$, then the conditions $g(x\_1)=\ln a$ and $x\_1\in(0,1)$ imply, successively, the following: $\ln x\_1\to-\infty$, $x\_1\to0$, $-\ln x\_1=\ln a+o(1)$, and $x\_1\sim1/a$. 
Also, again if $a\to\infty$, then the conditions $g(x\_2)=\ln a$ and $x\_2>1$ imply $x\_2^2\sim\ln a$ and hence $x\_2\sim\sqrt{\ln a}$. So, if $a\to\infty$, then $x\_1x\_2\sim\frac1a\,\sqrt{\ln a}$, while the proposed lower and upper bounds on $x\_1x\_2$ are $\frac3{2a+1}\sim\frac3{2a}$ and $\frac{\ln a}{a-1}\sim\frac{\ln a}a$. 


By tweaking the asymptotic expressions $1/a$ and $\sqrt{\ln a}$ for $x\_1$ and $x\_2$ as $a\to\infty$, we obtain lower and upper bounds on $x\_1$ and $x\_2$, useful for large enough $a$, as follows. 


The conditions $g(x\_1)=\ln a$ and $x\_1\in(0,1)$ yield $x\_1^2-x\_1<0$ and hence $\ln a=g(x\_1)<-\ln x\_1$, so that we get an upper bound on $x\_1$: 
\begin{equation\*}
 x\_1<X\_1^{up}:=X\_1^{up}(a):=\tfrac1a. 
\end{equation\*}
Therefore and because $x^2-x$ is decreasing in $x\in(0,1/2]$, we have $x\_1^2-x\_1>\frac1{a^2}-\frac1a\ge\frac1{2^2}-\frac12=-\frac14$ if $a\ge2$, whence $\ln a=g(x\_1)>-\frac14-\ln x\_1$, so that we get a lower bound on $x\_1$: 
\begin{equation\*}
 x\_1>X\_1^{lo}:=X\_1^{lo}(a):=\tfrac1{e^{1/4}a}\quad\text{for}\quad a\ge2. 
\end{equation\*}


The conditions $g(x\_2)=\ln a$ and $x\_2>1$ yield $x\_2^2=x\_2+\ln x\_2+\ln a\ge 1+\ln a$, so that we get a lower bound on $x\_2$: 
\begin{equation\*}
 x\_2>X\_2^{lo}:=X\_2^{lo}(a):=\sqrt{1+\ln a}. 
\end{equation\*}
Therefore and because $h(x):=\frac{x+\ln x}{x^2}$ is decreasing in $x\ge1$ (from $1$ to $0$), we have $h(x\_2)<h(\sqrt{1+\ln a})\le h(\sqrt{1+\ln 17})$ if $a\ge17$. So, $x\_2^2=h(x\_2)x\_2^2+\ln a<h(\sqrt{1+\ln 17})x\_2^2+\ln a$, and 
we get an upper bound on $x\_2$: 
\begin{equation\*}
 x\_2<X\_2^{up}:=X\_2^{up}(a):=\frac{\sqrt{\ln a}}c \quad\text{for}\quad a\ge17, 
\end{equation\*}
where $c:=\sqrt{1-h(\sqrt{1+\ln 17})}\approx0.560$. 


Consider now the ratio 
\begin{equation\*}
 R^{up}(a):=X\_1^{up}X\_2^{up}\Big/\frac{\ln a}{a-1}=\frac{a-1}{ca\sqrt{\ln a}}. 
\end{equation\*}
It is easy to see that $R^{up}(a)$ decreases in $a>4$, and $R^{up}(17)\approx0.9979<1$. So, 
\begin{equation\*}
 x\_1x\_2<X\_1^{up}X\_2^{up}<\frac{\ln a}{a-1} \quad\text{for}\quad a\ge17.
\end{equation\*}


Similarly, the ratio 
\begin{equation\*}
 R^{lo}(a):=X\_1^{lo}X\_2^{lo}\Big/\frac3{2a+1}=\frac{(2a+1)\sqrt{1+\ln a}}{3e^{1/4}a} 
\end{equation\*}
increases in $a>1$, and $R^{lo}(11)\approx1.00056>1$. So, 
\begin{equation\*}
 x\_1x\_2>X\_1^{lo}X\_2^{lo}>\frac3{2a+1} \quad\text{for}\quad a\ge11.
\end{equation\*}




---


**Small values of $a$.** Here we consider somewhat small values of  

\begin{equation\*}
 \ep:=a-1>0.
\end{equation\*}
Then Maclaurin expansions of the roots in powers of $\sqrt\ep$ are as follows: 
\begin{equation\*}
 x\_i\approx1+\sum\_{j=1}^4c\_j\,(-\sqrt\ep)^{ij},
\end{equation\*}
where $i\in\{1,2\}$ and 
\begin{equation\*}
 (c\_1,\dots,c\_4):=\Big(\sqrt{\frac{2}{3}},\frac{2}{27},-\frac{277}{486 \sqrt{6}},-\frac{893}{32805}\Big). 
\end{equation\*}
One does not have to check this, as we are just going to tweak these expansions a bit to obtain true lower and upper bounds on $x\_1$ and $x\_2$ for small enough $\ep=a-1>0$. 


Indeed, let 
\begin{align}
x\_1^{lo}&:=x\_1^{lo}(\ep):=1+\sum\_{j=1}^3c\_j\,(-\sqrt\ep)^{j}-\frac8{100}\ep^2, \\ 
x\_2^{lo}&:=x\_2^{lo}(\ep):=1+\sum\_{j=1}^4c\_j\,(\sqrt\ep)^{j}, \\ 
x\_1^{up}&:=x\_1^{up}(\ep):=1+\sum\_{j=1}^4c\_j\,(-\sqrt\ep)^{j}, \\ 
x\_2^{up}&:=x\_2^{up}(\ep):=1+\sum\_{j=1}^3c\_j\,(\sqrt\ep)^{j}+\frac1{20}\ep^2. 
\end{align}
The derivatives of $d\_i^{lo}(\ep):=g(x\_i^{lo})-\ln a=g(x\_i^{lo})-\ln(1+\ep)$ and $d\_i^{up}(\ep):=g(x\_i^{up})-\ln a=g(x\_i^{up})-\ln(1+\ep)$ are rational functions of $\sqrt\ep$, and monotonicity patterns of these functions can be quickly determined algorithmically, using a computer algebra program. E.g., $(d\_1^{lo})'(\ep)>0$ for $\ep\in(0,\frac{175}{1000}]$ and hence $d\_1^{lo}(\ep)$ increases in $\ep\in(0,\frac{175}{1000}]$, from $d\_1^{lo}(0)=0$. So, on $(0,\frac{175}{1000}]$ we have $d\_1^{lo}>0$, that is, $g(x\_1^{lo})>\ln a=g(x\_1)$; since $g$ is decreasing on $(0,1]$, it follows that 
\begin{equation\*}
 x\_1>x\_1^{lo}\quad\text{for}\quad \ep\in(0,\tfrac{175}{1000}]. 
\end{equation\*}
Similarly we get 
\begin{align}
 x\_2&>x\_2^{lo}\quad\text{for}\quad \ep\in(0,\tfrac{175}{1000}], \\ 
 x\_1&<x\_1^{up}\quad\text{for}\quad \ep\in(0,\tfrac{127}{1000}], \\ 
 x\_2&<x\_2^{up}\quad\text{for}\quad \ep\in(0,\tfrac{127}{1000}]. 
\end{align}


Further, $x\_1^{lo}x\_2^{lo}-\frac3{2a+1}$ and the derivative of $x\_1^{lo}x\_2^{lo}-\frac{\ln a}{a-1}$ are rational functions of $\sqrt\ep$, which allows one to similarly use a computer algebra program to quickly show that 
\begin{equation\*}
 x\_1x\_2<x\_1^{up}x\_2^{up}<\frac{\ln a}{a-1} \quad\text{for}\quad a\in(1,1+\tfrac{175}{1000}],
\end{equation\*}
\begin{equation\*}
 x\_1x\_2>x\_1^{lo}x\_2^{lo}>\frac3{2a+1} \quad\text{for}\quad a\in(1,1+\tfrac{127}{1000}]. 
\end{equation\*}




---


It remains to consider 


**Intermediate values of a.** In view of the above considerations of large and small values of $a$, it remains to prove the lower bound $\frac3{2a+1}$ on $x\_1x\_2$ for $a\in[1+\tfrac{175}{1000},11]$ and the lower bound $\frac{\ln a}{a-1}$ on $x\_1x\_2$ for $a\in[1+\tfrac{127}{1000},17]$. 
This is done by the mentioned interval method. 
Note that the monotonicity pattern of the function $g$ allows one to approximate the values of $x\_1$ and $x\_2$ with any prescribed degree of accuracy, for each $a>1$. However, here it is slightly more convenient to use the explicit (albeit very complicated) equivalent form of the inequalities in question, given in the last display in the OP: 
\begin{equation\*}
 \frac3{2e^{g(F(t))}+1}< t F(t)^2<\frac{g(F(t))}{e^{g(F(t))}-1}, \tag{1}
\end{equation\*}
where 
\begin{equation\*}
 F(t):=\frac{t-1+\sqrt{(t-1)^2+4\ln{t}\cdot(t^2-1)}}{2(t^2-1)}=x\_1\in(0,1) 
\end{equation\*}
and $t:=t(a):=x\_2/x\_1>1$. (The OP used the symbol $f$ to denote both $F$ and $g(x)-\ln a$.) 


Note that $x\_1$ is decreasing and $x\_2$ is increasing in $a$, so that $t$ is increasing in $a$. So, for $a\in[1+\tfrac{127}{1000},17]$ we have $t\in[t(1+\tfrac{127}{1000}),t(17)]\subseteq[t^{small},t^{large}]$, 
where $t^{small}:=x\_2^{up}(\tfrac{127}{1000})/x\_1^{lo}(\tfrac{127}{1000})\approx1.7699$ and $t^{large}:=X\_2^{up}(17)/X\_1^{lo}(17)=53.974\ldots$. 


Since $t$ is increasing and $x\_1=F(t)$ is decreasing in $a$, it follows that $F(t)\in(0,1)$ is decreasing in $t>1$. Therefore and because $g$ is decreasing on $(0,1)$, $g(F(t))$ is increasing in $t$. Also, $\frac b{e^b-1}$ is decreasing in $b>0$. So, 
\begin{equation\*}
 \frac{g(F(t))}{e^{g(F(t))}-1} 
\end{equation\*}
is decreasing in $t>1$. So, to prove the second inequality in (1) for $a\in[1+\tfrac{127}{1000},17]$, it suffices to show that 
\begin{equation\*}
 t\_{j+1} F(t\_j)^2<\frac{g(F(t\_{j+1}))}{e^{g(F(t\_{j+1}))}-1} \tag{2}
\end{equation\*}
for all integers $j\in\{0,\dots,53n\}$, where $n=300$, 
$t\_j:=\frac{176}{100}+\frac jn$, so that $t\_{53n}=53+\frac{176}{100}\ge53.974\ldots$. 
Indeed, then it will follow that 
\begin{equation\*}
 t F(t)^2\le t\_{j+1} F(t\_j)^2<\frac{g(F(t\_{j+1}))}{e^{g(F(t\_{j+1}))}-1}
 \le\frac{g(F(t))}{e^{g(F(t))}-1}
\end{equation\*}
for all $j\in\{0,\dots,53n\}$ and all $t\in[\frac{176}{100}+\frac jn,\frac{176}{100}+\frac{j+1}n]$, so that we will indeed have the second inequality in (1) for all $t\in[t(1+\tfrac{127}{1000}),t(17)]$. 
The proof of (2) is done by straightforward calculation (taking about 14 sec with Mathematica). 


The proof of the first inequality in (1) (for $t\in[t(1+\tfrac{175}{1000}),t(11)]$) is quite similar (and a bit easier); it takes about 6 sec with Mathematica. 


The proof is now complete. 


