An Inequality Concerning the Roots of a transcendental equation

Question

Let $f(x)=x(1-\ln x)$ and $a,b,c \in \mathbb{R}$. If $a,b$ are the distinct roots of $f(1+({\rm e}-2)x)=c$,while e is the base of the natural logarithm,prove that: $$0<c<1<a+b+2c<2$$

My Attempt:Firstly it is easy to prove $0<c<1$:while$$f'(x)=-\ln x$$ Thus, the function is monotonically increasing on (0,1) and monotonically decreasing on (1,+∞). Additionally, $f(e)=0$, and for $x>e$, we have $f(x)<0$.We only need to calculate the range of values of $ f(x) $ on the interval (0, e). Since the limit of $ f(x) $ as $ x $ approaches 0 is 0, the range of values is $(0, 1]$. Therefore, $$0 < c < 1$$ Then consider $1<a+b+2c<2$.I have no idea about the left side,but to the right side,maybe we can firstly prove $a+b+c<1$ and it has been tested by software,but I still have no idea.

Answer 1

Here is a proof for $1 < a + b + 2c$.

Let $a_1 := 1 + (\mathrm{e} - 2)a$ and $b_1 := 1 + (\mathrm{e} - 2)b$. Then $a_1, b_1 > 0$ are two distinct real roots of $f(x) = c$ given that $0 < c < 1$. We need to prove that $$1 < \frac{a_1 + b_1 - 2}{\mathrm{e} - 2} + 2c. \tag{1}$$

WLOG, assume that $a_1 < b_1$. Since $f(x)$ is strictly increasing on $(0, 1)$ and strictly decreasing on $(1, \infty)$, and $\lim_{x\to 0^{+}} f(x) = 0$, and $f(x) \le 0$ for all $x \ge \mathrm{e}$, we have $0 < a_1 < 1 < b_1 < \mathrm{e}$.

We claim that $a_1 + b_1 \ge 2$. Indeed, letting $$g(a_1) := f(2- a_1) - f(b_1) = (2-a_1)[1 - \ln(2 - a_1)] - a_1(1 - \ln a_1),$$ we have $g'(a_1) = \ln(2 - a_1) + \ln a_1 = \ln(1 - (1-a_1)^2) < 0$ and $g(1) = 0$, which results in $g(a_1) \ge 0$ and thus $2 - a_1 \le b_1$. The claim is proved.

If $c > 1/2$, using $a_1+b_1\ge 2$, (1) is clearly true.

If $0 < c \le 1/2$, it suffices to prove that $$1 \le \frac{b_1 - 2}{\mathrm{e} - 2} + 2c, $$ or $$b_1 \ge \mathrm{e} - 2(\mathrm{e}-2)c.$$

Note that $\mathrm{e} - 2(\mathrm{e}-2)c > 1$. Let $$h(c) := f(\mathrm{e} - 2(\mathrm{e}-2)c) - f(b_1) = f(\mathrm{e} - 2(\mathrm{e}-2)c) - c.$$ We have $h''(c) = - \frac{4(\mathrm{e}-2)^2}{\mathrm{e} - 2(\mathrm{e}-2)c} < 0$ (concave) and $h(0) = 0$ and $h(1/2) > 0 $, which results in $h(c) \ge 0$ on $(0, 1/2]$ and thus $b_1 \ge \mathrm{e} - 2(\mathrm{e}-2)c$. (1) is true.

We are done.

Answer 2

Here is a proof using the implicit differentiation.

Remarks. I tried this approach yesterday but was stuck at the proof of $F'(c) > 0$. Now I can proceed. I learned this approach some years ago from @G.Kós's answer for this question. By the way, I noticed that @Macavity has tried this way as well.

Let $a_1 := 1 + (\mathrm{e} - 2)a$ and $b_1 := 1 + (\mathrm{e} - 2)b$. Then $a_1, b_1 > 0$ are two distinct real roots of $f(x) = c$ given that $0 < c < 1$. We need to prove that $$1 < \frac{a_1 + b_1 - 2}{\mathrm{e} - 2} + 2c < 2. \tag{1}$$

WLOG, assume that $a_1 < b_1$. Since $f(x)$ is strictly increasing on $(0, 1)$ and strictly decreasing on $(1, \infty)$, and $\lim_{x\to 0^{+}} f(x) = 0$, and $f(x) \le 0$ for all $x \ge \mathrm{e}$, we have $0 < a_1 < 1 < b_1 < \mathrm{e}$.

For $0 < c < 1$, the two real roots $a_1 = a_1(c)$ and $b_1 = b_1(c)$ are functions of $c$ satisfying $$a_1(c) - a_1(c) \ln a_1(c) = b_1(c) - b_1(c) \ln b_1(c) = c.$$ By implicit differentiation, we have $$a_1'(c) - a_1'(c)\ln a_1(c) - a_1(c) \frac{1}{a_1(c)}a_1'(c) = 1, $$ and $$b_1'(c) - b_1'(c)\ln b_1(c) - b_1(c) \frac{1}{b_1(c)}b_1'(c) = 1$$ which result in $$a_1'(c) = -\frac{1}{\ln a_1(c)}, \quad b_1'(c) = - \frac{1}{\ln b_1(c)}.$$

Now, let $$F(c) := \frac{a_1(c) + b_1(c) - 2}{\mathrm{e} - 2} + 2c.$$ We have $$ F'(c) = \frac{a_1'(c) + b_1'(c)}{\mathrm{e} - 2} + 2 = \frac{-\frac{1}{\ln a_1(c)} - \frac{1}{\ln b_1(c)}}{\mathrm{e} - 2} + 2 > 0. \tag{2} $$ The proof of (2) is given at the end.

Also, $\lim_{c\to 0^{+}} F(c) = 1$ and $\lim_{c\to 1^{-}} F(c) = 2$. Thus, we have $1 < F(c) < 2$.

We are done.

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Proof of (2).

If $0 < c \le \frac13$, we have $f(7/3) = \frac73 - \frac73 \ln \frac73 > \frac13 \ge c = f(b_1(c))$ which results in $b_1(c) \ge \frac73$. Thus, we have $$\frac{-\frac{1}{\ln a_1(c)} - \frac{1}{\ln b_1(c)}}{\mathrm{e} - 2} + 2 > \frac{ - \frac{1}{\ln b_1(c)}}{\mathrm{e} - 2} + 2 \ge \frac{ - \frac{1}{\ln \frac73}}{\mathrm{e} - 2} + 2 > 0.$$

If $c > \frac13$, we have $f(1/10) = \frac{1}{10} - \frac{1}{10}\ln \frac{1}{10} < \frac13 < c = f(a_1(c))$ which results in $a_1(c) \ge \frac{1}{10}$. Using $a_1(c) + b_1(c) \ge 2$ (the proof is given in my previous answer), we have $$\frac{-\frac{1}{\ln a_1(c)} - \frac{1}{\ln b_1(c)}}{\mathrm{e} - 2} + 2 \ge \frac{-\frac{1}{\ln a_1(c)} - \frac{1}{\ln [2 - a_1(c)]}}{\mathrm{e} - 2} + 2 \ge \frac{-\frac43}{\mathrm{e} - 2} + 2 > 0$$ where we use $-\frac{1}{\ln u} - \frac{1}{\ln (2-u)} \ge - \frac43$ for all $1/10 \le u < 1$ (the proof is given at the end).

We are done.

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Proof of $-\frac{1}{\ln u} - \frac{1}{\ln (2-u)} \ge - \frac43$ for all $1/10 \le u < 1$.

After clearing the denominators, it suffices to prove that $$G(u) := -4\ln u \ln (2 - u) + 3\ln u + 3\ln (2 - u) \ge 0.$$

Let $H(u) := G'(u)u(2-u)$. We have $$H''(u) = \frac{8(1-u)}{u(2-u)} > 0.$$ Also, $H(1) = 0$ and $H(1/10) < 0$. Thus, $H(u) \le 0$ on $[-1/10, 1)$. Thus, $G'(u) \le 0$ on $[-1/10, 1)$. Also, $G(1) = 0$. Thus, $G(u) \ge 0$ on $[-1/10, 1)$.

We are done.