取值范围求解

Analyzing the inequality

I'm working through the inequality $x e^x - a \ln x - a x - a \geq 0$ for all $x > 0$. Now, I need to find the range of $a$ that satisfies this condition.

Identifying the inequality

I’m determining values for $a$ such that $x e^x - a \ln x - a x - a \geq 0$ for all $x > 0$.

Optimizing $a$

I’m optimizing $a$ to ensure the inequality holds for all $x > 0$. This involves finding the minimum value of $\frac{ x e^x }{ \ln x + x + 1 }$ to determine the threshold for $a$.

Pinpointing constraints

I’m noting that for $f(x)$ to remain non-negative for $x > 0$, we need to find the minimal value of $f(x)$ as a function of $a$.

Breaking down the problem

I’m setting $f(x) \geq 0$ and factoring out $a$ in the expression. This leads me to explore the relationship between $a$, $x$, and $\ln x + x + 1$.

Assessing the behavior

I’m noticing how $\ln x$ changes sign as $x$ crosses zero, and how $\ln x + x + 1$ varies around $x = 0$, $x = 1$, and as $x$ approaches infinity.

Crafting the approach

I'm thinking through how to find the minimal value of $f(x)$ by deriving $f'(x)$ and pinpointing the critical points.

Breaking down $f(x)$ into derivatives

I’m piecing together $f(x)$ by splitting each term. Simplifying derivatives, I end up with $f'(x) = e^x ( x + 1 ) - \frac{a}{x} - a$, though it's a bit tricky to manage.

Identifying function behavior

OK, let me see. As $x$ approaches 0 from the right, $f(x)$ heads to infinity due to $-a \ln x$'s dominance.

Tracing the critical points

I'm piecing together the critical points of $f(x)$ by setting $f'(x) = 0$ to pinpoint where a minimum might logically occur.

Isolating $a$ and finding critical points

I’m working through $e^x ( x + 1 ) = a \left( \frac{1}{x} + 1 \right)$ and finding that $a = e^x x$ at critical points.

Determining critical points

OK, I’m noting that for $x > 0$, the derivative $f'(x)$ equals zero at $a = e^x x$. This leads to the critical points.

Substituting and simplifying

OK, let me see. I've substituted $a = x e^x$ into $f(x)$, but the simplification seems a bit off. I’m working through the expression to clarify it.

Breaking down the equation

I’m noting that $a = x e^x$ and simplifying $f(x)$ by substituting and combining terms. This highlights the algebraic progression, leading to a more concise form.

Breaking down the function

I started with expressions for $a$ and $a \ln x$, simplified them to $f(x)$, and reduced it to $- x e^x ( \ln x + x )$.

Breaking down the function

I’m working out the function $f(x) = - x e^x ( \ln x + x )$ and rearranging it as $g(x) = \ln x + x$. Next, I’ll figure out its critical points.

Working through the steps

I’m calculating $g'(x)$ for $g(x) = \ln x + x$. It’s positive for $x > 0$, so $g(x)$ is increasing in that range.

Solving a transcendental equation

I'm working on finding $x$ for $\ln x + x = 0$. This involves potentially iterative methods or specialized techniques due to its transcendental nature.

Determining critical points

I calculated the values of $g(x)$ at different points and found that $x \approx 0.5671$ is the solution of $\ln x + x = 0$, where $g(x)$ changes sign.

Evaluating minimal value

I discovered that $f(x) = - x e^x ( \ln x + x )$ can be negative when $x > 0.5671$. This indicates that $f(x)$ might not always be non-negative, prompting a reassessment of the initial assumption.

Evaluating zero crossing

I determined that $g(x)$ crosses zero at $x \approx 0.5671$ by calculating $g(x)$ at $x = e^{-1}$ and $x = 1$.

Calculating $f(x)$ value

I computed $f(1)$ and found it to be negative, specifically $-e \approx -2.71828$.

Determining behavior

I'm mapping out the behavior of $f(x)$ around $x = 0.5671$, noting it decreases from $+\infty$ to $-e$ before becoming negative.

Finding the threshold

I’m looking to determine the smallest possible value $a$ that ensures $f(x) \geq 0$ for all $x > 0$, given that $f(x)$ sometimes turns negative for certain $x$.

Testing specific values

I’m examining the inequality for $a \leq 0$. With $a = 0$, $f(x) \geq 0$ for $x > 0$. For $a < 0$, $f(x)$ can be negative due to the $\ln x$ term.

Evaluating non-negativity

Initially, I’m working through if $a \leq 0$ ensures $f(x) \geq 0$ for all $x > 0$, as the terms are non-negative with $-a \geq 0$.

Investigating function behavior

I’m analyzing the constraints on $a$ in the function $f(x)$ to ensure $-a(\ln x + x + 1) \geq 0$. This leads me to conclude that $a \leq 0$ is necessary for non-negativity.

Breaking down the equation

I’m examining the term $-a (\ln x + x + 1) \leq 0$ to find the range of $a$. This involves locating the minimal value of $f(x)$.