inequality related to roots of $(x-1)\log(x)=m$

Question

Let $f(x) := (x-1)\log {x}$. Suppose $f(x_1)=f(x_2)=m$ for some $0<x_1<x_2$. Show that $\frac{9}{5}+\log{(1+m)}<x_1+x_2<2+\frac{m}{2}$.

If we apply Hermite-Hadamard inequality, it's easy to show $2<x_1+x_2$, but it's not strong enough. I also tried to replace $m$ with $f(x_1)$, then it became super complicated. There should be some easy way to simplify $x_1, x_2$

Answer 1

The upper bound of $2+\frac m2$ is proven.

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For $0<x_1<1$ and $x_2>1$, the equations $$(x_1-1)\ln x_1=m\implies x_1=1+\frac m{\ln x_1}\\(x_2-1)\ln x_2=m\implies x_2=1+\frac m{\ln x_2}$$ can be added to give $x_1+x_2=2+m\left(\frac1{\ln x_1}+\frac1{\ln x_2}\right)$ and for the upper bound of $2+\frac m2$ it suffices to prove that $$\frac1{\ln x_1}+\frac1{\ln x_2}<\frac12\impliedby x_2>\exp\left(\frac{2\ln x_1}{\ln x_1-2}\right).$$ The smallest value of $x_2$ will yield $\frac1{\ln x_1}+\frac1{\ln x_2}$ closest to $\frac12$, so in the worst case, we suppose equality of the above. Substituting this equation into the one for $x_2$, we get $$(x_2-1)\ln x_2>\left[\exp\left(\frac{2\ln x_1}{\ln x_1-2}\right)-1\right]\frac{2\ln x_1}{\ln x_1-2}$$ and in turn it suffices to show that $$\left[\exp\left(\frac{2\ln x_1}{\ln x_1-2}\right)-1\right]\frac{2\ln x_1}{\ln x_1-2}\ne (x_1-1)\ln x_1$$ for $0<x_1<1$ as we want to prove that the LHS is always smaller than the RHS (e.g. at $x_1=\frac12$ this is true so if there is an interval where the LHS is greater than there must be a point where the two curves meet due to continuity).

Thus, suppose that they are equal. Rearranging gives $$f(x_1)=2x_1^{\frac2{\ln x_1-2}}+2x_1-4-(x_1-1)\ln x_1=0$$ Now $$f'(x_1)=-\frac8{x_1(\ln x_1-2)^2}x_1^{\frac2{\ln x_1-2}}+\frac1{x_1}-\ln x_1+1$$ and letting $u=2-\ln x_1$ results in $$f'(u)=\frac{1}{\exp\left(2-u\right)}-\frac{8}{u^2}\exp\left(u-\frac{4}{u}\right)+u-1$$ for $u\in(2,\infty)$. This is now computationally viable and W|A shows that indeed $f'(u)>0$ for $u\in(2,\infty)$, and $f'(u)=0$ only at $u=2$. Hence there does not exist $x_1$ such that $f(x_1)=0$, so the upper bound holds for all $x_1,x_2$ in their respective domains.

Answer 2

Alternative solution for $x_1 + x_2 < 2 + m/2$.

Clearly, $0 < x_1 < 1 < x_2$. We need to prove that $x_2 < 2 + m/2 - x_1$.

Note that $f(x)$ is strictly increasing on $(1, \infty)$ since $f'(x) = \ln x + (x-1)/x > 0$. It suffices to prove that $f(x_2) < f(2 + m/2 - x_1)$. Using $f(x_2) = f(x_1)$ and $m = (x_1 - 1)\ln x_1$, it suffices to prove that $$(x_1 - 1)\ln x_1 < (1 + 2^{-1}(x_1 - 1)\ln x_1 - x_1)\cdot \ln(2 + 2^{-1}(x_1 - 1)\ln x_1 - x_1), $$ or $$(x_1 - 1)\ln x_1 < (1 - x_1)(1 - 2^{-1}\ln x_1) \cdot \ln(2 + 2^{-1}(x_1 - 1)\ln x_1 - x_1),$$ or $$-\ln x_1 < (1 - 2^{-1}\ln x_1) \cdot \ln(2 + 2^{-1}(x_1 - 1)\ln x_1 - x_1),$$ or $$F(x_1) := \ln(2 + 2^{-1}(x_1 - 1)\ln x_1 - x_1) - \frac{-\ln x_1}{1 - 2^{-1}\ln x_1} > 0.$$

Since $F(1) = 0$. It suffices to prove that $F'(x_1) < 0$ for all $x_1 \in (0, 1)$, or equivalently, $$g(x_1) := x_1 \ln^3 x_1 - 5x_1 \ln^2 x_1 - \ln^2 x_1 + 12x_1\ln x_1 - 12x_1 + 12 < 0.$$ Since $g(1) = 0$, it suffices to prove that $g'(x_1) > 0$ for all $x_1 \in (0, 1)$, or equivalently, $$h(x_1) := -x_1\ln^2 x_1 + 2x_1\ln x_1 - 2x_1 + 2 > 0.$$ We have $h'(x_1) = -\ln^2 x_1 < 0$. Also, we have $h(1) = 0$. Thus, $h(x_1) > 0$ on $(0, 1)$.

We are done.

Answer 3

For the left inequality.

Remark. The approach is similar to my answer for this question.

Note that $f(x)$ is strictly decreasing on $(0, 1)$ and strictly increasing on $(1, \infty)$, and $\lim_{x\to 0^{+}} f(x) = \infty$, and $\lim_{x\to \infty} f(x) = \infty$, and $f(1) = 0$. Thus, we have $m > 0$ and $0 < x_1 < 1 < x_2$.

We split into two cases.

Case 1. $m \ge 3$

It suffices to prove that $\frac95 + \ln(1+m) \le x_2$ or $$f\left(\frac95 + \ln(1+m) \right) \le f(x_2) = m,$$ or $$\left(\frac95 + \ln(1+m) - 1\right)\ln \left(\frac95 + \ln(1+m)\right) \le m$$ which is true (not difficult by calculus and using $\ln(1 + m) \ge \frac{2m}{2+m}$).

Case 2. $0 < m < 3$

Let $$g(x) := \frac{3(x-1)^2(19x+11)}{(x+2)(-x^2 + 26x + 5)}.$$ We have $$f(x) - g(x) \left\{\begin{array}{ll} > 0 & 0 < x < 1\\ = 0 & x = 1\\ < 0 & 1 < x < 20. \end{array} \right. \tag{1}$$

Note that $g(x)$ is strictly decreasing on $(0, 1)$ and strictly increasing on $(1, 20)$, and $g(1) = 0$, and $g(0) = 33/10 > 3$, and $g(20) > 3$. Thus, $g(x) = m$ has exactly two distinct real roots $0 < x_3 < 1 < x_4 < 20$. Also, from (1), we have $x_3 < x_1$ and $x_4 < x_2$. Thus, we have $x_1 + x_2 > x_3 + x_4$. It suffices to prove that $$x_3 + x_4 \ge \frac95 + \ln(1 + m). \tag{2}$$

$x_3, x_4$ are two distinct real roots of the cubic equation $$3(x-1)^2(19x+11) = m (x+2)(-x^2 + 26x + 5), $$ that is $$(m + 57)x^3 + (-24m - 81)x^2 + (-57m - 9)x - 10m + 33 = 0. \tag{3}$$ Denote the third root of (3) by $x_5$. By Vieta's theorem, we have $x_3 + x_4 + x_5 = \frac{24m + 81}{m + 57} > 0, x_3x_4 + x_4x_5 + x_5x_3 = - \frac{57m + 9}{m + 57}, x_3 x_4 x_5 = \frac{10m - 33}{m+57} < 0$. Clearly, we have $x_5 < 0$.

Consider the cubic equation $F(u) = 0$ where \begin{align*} F(u) &:= (u - x_3 - x_4)(u - x_4 - x_5)(u - x_5 - x_3)\\ &= u^3 - \frac{2(24m + 81)}{m+57}u^2 + \frac{3(173m^2 + 210m + 2016)}{(m+57)^2}u \\ &\qquad + \frac{2(689m^2 + 2685m - 576)}{(m+57)^2}. \end{align*} Then, $x_3 + x_4, x_4+x_5, x_5 + x_3$ are the three distinct real roots of $F(u) = 0$. Since $x_3 + x_4$ is the largest real root of $F(u) = 0$, we have $F(u) \ge 0$ for all $u \ge x_3 + x_4$. Thus, to prove (2), it suffices to prove that $$F\left(\frac95 + \ln(1 + m)\right) < 0 \tag{4}$$ which is true. The proof of (4) is given at the end.
(Note: If $x_3 + x_4 < \frac95 + \ln(1 + m)$, then we have $F(\frac95 + \ln(1 + m)) \ge 0$.)

We are done.

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Proof of (4).

It suffices to prove that \begin{align*} &(125m^2+14250m+406125)Q^3 + (-5325m^2-285300m+1038825)Q^2\\ &\qquad + (44490m^2-1086840m+548235)Q\\ &\qquad +270314m^2-277584m-154449 < 0 \tag{A1} \end{align*} where $Q := \ln(1 + m)$. (A1) is true. However, my proof is complicated. Omitted here.

Answer 4

My second proof for the left inequality.

Remark. The trouble is $m \approx 1$ for which $x_1 + x_2 - \frac95 - \ln(1+m)$ is small. For example, when $m=1$, $x_1 + x_2 - \frac95 - \ln(1+m) \approx 0.006$.

Note that $f(x)$ is strictly decreasing on $(0, 1)$ and strictly increasing on $(1, \infty)$, and $\lim_{x\to 0^{+}} f(x) = \infty$, and $\lim_{x\to \infty} f(x) = \infty$, and $f(1) = 0$. Thus, we have $m > 0$ and $0 < x_1 < 1 < x_2$. We need to prove that $$\frac95 + \ln(1 + m) < x_1 + x_2. \tag{1}$$

We split into three cases.

• Case 1. $0 < m \le \frac23$

Let $$g(x) := \frac{12(x-1)^2}{-x^2 + 8x + 5}.$$

We have $$f(x) - g(x) \left\{\begin{array}{ll} > 0 & 0 < x < 1\\ = 0 & x = 1\\ < 0 & 1 < x < 3. \end{array} \right. \tag{2}$$

Note that $g(x)$ is strictly decreasing on $(0, 1)$ and strictly increasing on $(1, 3)$, and $g(1) = 0$, and $g(0) = 12/5 > 2/3$, and $g(3) = 12/5 > 2/3$. Thus, $g(x) = m$ has exactly two distinct real roots $0 < x_3 < 1 < x_4 < 3$. Also, from (2), we have $x_3 < x_1$ and $x_4 < x_2$. Thus, we have $x_1 + x_2 > x_3 + x_4$. It suffices to prove that $$\frac95 + \ln(1 + m) \le x_3 + x_4. \tag{3}$$

$x_3, x_4$ are two distinct real roots of the quadratic equation $12(x-1)^2 = m(-x^2 + 8x + 5)$ that is $$(m+12)x^2 + (-8m-24)x - 5m+12 = 0. $$ By Vieta's theorem, we have $x_3 + x_4 = \frac{8m + 24}{m+12}$. To prove (3), it suffices to prove that $$\frac95 + \ln(1 + m) \le \frac{8m + 24}{m+12}$$ which is true (easy by calculus).

• Case 2. $2/3 < m < 103/50$

We claim that $x_1 > \frac{8}{31} - \frac{7}{30}(m - 1) > 0$. It suffices to prove that $$f(x_1) = m < f\left(\frac{8}{31} - \frac{7}{30}(m - 1)\right),$$ or $$m < - \left(\frac{473}{930} + \frac{7}{30}m\right)\ln\left(\frac{457}{930} - \frac{7}{30}m\right)$$ which is true (not difficult by calculus). The claim is proved.

To prove (1), it suffices to prove that $$\frac95 + \ln(1 + m) \le \frac{8}{31} - \frac{7}{30}(m - 1) + x_2, $$ or $$\frac{1217}{930} + \frac{7}{30}m + \ln(1 + m) \le x_2.$$ It suffices to prove that $$f\left(\frac{1217}{930} + \frac{7}{30}m + \ln(1 + m)\right) \le f(x_2) = m,$$ or $$\left(\frac{287}{930} + \frac{7}{30}m + \ln(1 + m)\right)\ln\left(\frac{1217}{930} + \frac{7}{30}m + \ln(1 + m)\right) \le m, $$ or $$H(m) := \frac{m}{\frac{287}{930} + \frac{7}{30}m + \ln(1 + m)} - \ln\left(\frac{1217}{930} + \frac{7}{30}m + \ln(1 + m)\right) \ge 0.$$ It is not difficult to prove that $H'(m) > 0$ (using $\ln(1 + m) \ge \frac{9}{13} + \frac{2(m-1)}{3+m}$ for all $m\ge 2/3$). Also, $H(2/3) > 0$. Thus, $H(m) \ge 0$ on $[2/3, 103/50]$.

• Case 3. $m \ge \frac{103}{50}$

To prove (1), it suffices to prove that $\frac95 + \ln(1+m) \le x_2$. It suffices to prove that $$f\left(\frac95 + \ln(1+m) \right) \le f(x_2) = m,$$ or $$\left(\frac95 + \ln(1+m) - 1\right)\ln \left(\frac95 + \ln(1+m)\right) \le m,$$ or $$G(m) := \frac{m}{4/5 + \ln(1 + m)} - \ln \left(\frac95 + \ln(1+m)\right) \ge 0.$$ It is not difficult to prove that $G'(m) > 0$ (using $\ln(1 + m) \ge \frac{2m}{2+m}$ for all $m \ge 0$). Also, $G(103/50) > 0$. Thus, $G(m) \ge 0$ on $[103/50, \infty)$.

We are done.