LaTeX Test Article

Just a test article to check if LaTeX formulas render properly.

Given that $e^{x_1}-ax_1=e^{x_2}-ax_2=0$, $x_1 \neq x_2$, we are to prove:

Define $f(x)=e^x-ax$. Since $f(x_1)=f(x_2)=0$, we have $x_1-\ln x_1=\ln a$ and $x_2-\ln x_2=\ln a$.

Define $g(x)=x-\ln x$. Thus, $g(x_1)=g(x_2)=\ln a$. Since $g'(x)=\frac{x-1}{x}$, $g(x)$ is strictly decreasing on $(0,1)$ and strictly increasing on $(1,+\infty)$. Without loss of generality, assume $0<x_1<1<x_2$.

Substituting $\ln a=x_1-\ln x_1$ into the target inequality, we transform it into:

Define $h(x)=\frac{x-4\ln x+2}{3}$. For $x \in (0,1)$, $h'(x)=\frac{x-4}{3x}<0$. Since $h(x)$ is strictly decreasing on $(0,1)$, $h(x_1)>h(1)=1$. Because $x_2>1$, $h(x_1)>1$, and $g(x)$ is strictly increasing on $(1,+\infty)$, the inequality is equivalent to:

Define $F(x)=g\left(\frac{x-4\ln x+2}{3}\right)-(x-\ln x)$ for $x \in (0,1)$. Computing the derivative yields:

where $M(x)=x^2+4x-2(2x+1)\ln x-5$.

For $M(x)$ on $(0,1)$:

Since $M''(x) \ge 0$, $M'(x)$ is strictly increasing. With $\lim_{x \to 1^-} M'(x)=0$, we have $M'(x)<0$. Thus, $M(x)$ is strictly decreasing. With $\lim_{x \to 1^-} M(x)=0$, we have $M(x)>0$.

Because $M(x)>0$ on $(0,1)$, it follows that $F'(x)<0$. $F(x)$ is strictly decreasing on $(0,1)$. Since $\lim_{x \to 1^-} F(x)=0$, $F(x)>0$ for all $x \in (0,1)$.

Therefore, the equivalent inequality holds:

Substituting $-4\ln x_1=4(\ln a-x_1)$: