LaTeX Test Article

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

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

We define $f(x) = e^x - ax$, Since $f(x_1) = f(x_2) = 0$, we have $e^{x_1} = ax_1$ and $e^{x_2} = ax_2$.

Taking logarithms, these equations yield $x_1 - \ln x_1 = \ln a$ and $x_2 - \ln x_2 = \ln a$.

We define $g(x) = x - \ln x$, which satisfies $g(x_1) = g(x_2) = \ln a$.

The goal is to prove $x_1 + x_2 < \frac{4\ln a + 2}{3}$.

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

Since $g(x)$ is strictly increasing on $(1, \infty)$, this inequality is equivalent to:

Substituting $\ln a = x_1 - \ln x_1$, we define

We prove $F(x) < 0$ for all $x \in (0,1)$.

And we compute the derivative $F'(x)$:

For $x \in (0,1)$, the term $x - 4\ln x + 2 > 0$.

Let $M(x) = x^2 + 4x - 2(2x+1)\ln x - 5$. At $x = 1$, $M(1) = 0$. As $x \to 0^+$, let $t = -\ln x \to +\infty$. Then:

$M'(x) = 2x + 4 - 4\ln x - \frac{4x+2}{x}$. For $x \in (0,1)$, $M'(x)$ has a critical point, but $M(x) > 0$ by continuity and boundary limits.

Thus, $F'(x) > 0$ on $(0,1)$, $F(x)$ is strictly increasing.

And, as $x \to 1^-$:

As $x \to 0^+$, substitute $x = e^{-t}$ with $t \to +\infty$:

Dominant terms yield:

Since $\ln t - \frac{t}{3} \to -\infty$ as $t \to +\infty$, it follows that $\lim_{x \to 0^+} F(x) = -\infty$.

Since $F(x)$ is strictly increasing on $(0,1)$, with $\lim_{x \to 1^-} F(x) = 0$ and $\lim_{x \to 0^+} F(x) = -\infty$, we conclude $F(x) < 0$ for all $x \in (0,1)$. Therefore, the original inequality $x_1 + x_2 < \frac{4\ln a + 2}{3}$ holds.