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:

$$ x_1 + x_2 < \frac{4\ln a + 2}{3}. $$

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:

$$ x_2 < \frac{1}{3}x_1 - \frac{4}{3}\ln x_1 + \frac{2}{3}. $$

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

$$ \ln a < g\left( \frac{1}{3}x_1 - \frac{4}{3}\ln x_1 + \frac{2}{3} \right). $$

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

$$ F(x) = \ln\left( \frac{1}{3}x - \frac{4}{3}\ln x + \frac{2}{3} \right) + \frac{2}{3}x + \frac{1}{3}\ln x - \frac{2}{3}. $$

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

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

$$ F'(x) = \frac{\frac{1}{3} - \frac{4}{3x}}{\frac{1}{3}x - \frac{4}{3}\ln x + \frac{2}{3}} + \frac{2}{3} + \frac{1}{3x}. $$

$$ = \frac{2\left(x^2 + 4x - 2(2x+1)\ln x - 5\right)}{3x\left(x - 4\ln x + 2\right)}. $$

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:

$$ M(x) = e^{-2t} + 4e^{-t} + 2(2e^{-t} + 1)t - 5 \geq 2t - 5 \to +\infty. $$

$ 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^- $:

$$ \lim_{x \to 1^-} F(x) = 0. $$

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

$$ F(x) = \ln\left( \frac{4}{3}t + o(t) \right) + \frac{2}{3}e^{-t} - \frac{t}{3} - \frac{2}{3}. $$

Dominant terms yield:

$$ \ln\left( \frac{4}{3}t \right) - \frac{t}{3} = \ln\left( \frac{4}{3} \right) + \ln t - \frac{t}{3}. $$

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.