Let $D:=\{(x, y)\mid x,y\in\mathbb R_+,x \ne y,x^y=y^x\}$. (Obvious that $x\ne1$ and $y\ne1$). And let $\alpha\le\beta$ be positive real numbers. Find
$$\inf_{(x,y)\in D}x^\alpha y^\beta.$$

[i]Proposed by Arkady Alt[/i]
1. **Understanding the Set \( D \)**:
   The set \( D \) is defined as:
   \[
   D := \{(x, y) \mid x, y \in \mathbb{R}_+, x \ne y, x^y = y^x\}
   \]
   This implies that \( x \) and \( y \) are positive real numbers, not equal to each other, and satisfy the equation \( x^y = y^x \).

2. **Analyzing the Equation \( x^y = y^x \)**:
   To understand the relationship between \( x \) and \( y \), we take the natural logarithm on both sides:
   \[
   \ln(x^y) = \ln(y^x) \implies y \ln x = x \ln y
   \]
   Rearranging this, we get:
   \[
   \frac{\ln x}{x} = \frac{\ln y}{y}
   \]
   Let \( f(t) = \frac{\ln t}{t} \). The function \( f(t) \) is known to have a maximum at \( t = e \) (where \( e \) is the base of the natural logarithm).

3. **Behavior of \( f(t) \)**:
   Since \( f(t) \) has a maximum at \( t = e \), for \( t \ne e \), \( f(t) \) is strictly decreasing for \( t > e \) and strictly increasing for \( t < e \). Therefore, if \( x \ne y \) and both \( x \) and \( y \) are not equal to \( e \), one must be greater than \( e \) and the other less than \( e \).

4. **Finding the Infimum**:
   We need to find:
   \[
   \inf_{(x,y) \in D} x^\alpha y^\beta
   \]
   Given \( \alpha \le \beta \), we consider the behavior of \( x^\alpha y^\beta \) under the constraint \( x^y = y^x \).

5. **Considering \( x > 1 \) and \( y > 1 \)**:
   Suppose \( x > 1 \) and \( y > 1 \). From the previous analysis, we know that \( x \ne y \) and \( x^y = y^x \). We need to show that \( x^y > e^e \).

6. **Proving \( x^y > e^e \)**:
   Since \( f(t) = \frac{\ln t}{t} \) has a maximum at \( t = e \), for \( x \ne y \) and \( x, y > 1 \), we have:
   \[
   x^y = y^x \implies x^y > e^e
   \]
   This is because \( x \) and \( y \) must be on opposite sides of \( e \) (one greater than \( e \) and one less than \( e \)), and the product \( x^y \) will be greater than \( e^e \).

7. **Conclusion**:
   Since \( x^y > e^e \) for all \( (x, y) \in D \), the infimum of \( x^\alpha y^\beta \) will be achieved when \( x \) and \( y \) are as close to \( e \) as possible, but still satisfying \( x \ne y \). Therefore, the infimum is:
   \[
   \inf_{(x,y) \in D} x^\alpha y^\beta = e^{\alpha + \beta}
   \]

The final answer is \( \boxed{ e^{\alpha + \beta} } \).