For all positive real numbers $x$ and $y$ let
\[f(x,y)=\min\left( x,\frac{y}{x^2+y^2}\right) \]
Show that there exist $x_0$ and $y_0$ such that $f(x, y)\le f(x_0, y_0)$ for all positive $x$ and $y$, and find $f(x_0,y_0)$.
1. We are given the function \( f(x,y) = \min\left( x, \frac{y}{x^2 + y^2} \right) \) and need to show that there exist \( x_0 \) and \( y_0 \) such that \( f(x, y) \le f(x_0, y_0) \) for all positive \( x \) and \( y \), and find \( f(x_0, y_0) \).

2. Let's consider the value of \( f \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \):
   \[
   f \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \min \left( \frac{1}{\sqrt{2}}, \frac{\frac{1}{\sqrt{2}}}{\left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2} \right)
   \]
   Simplifying the second term inside the minimum function:
   \[
   \left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2} + \frac{1}{2} = 1
   \]
   Therefore:
   \[
   \frac{\frac{1}{\sqrt{2}}}{1} = \frac{1}{\sqrt{2}}
   \]
   So:
   \[
   f \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \min \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}}
   \]

3. Now, we need to show that \( f(x, y) \le \frac{1}{\sqrt{2}} \) for all positive \( x \) and \( y \).

4. Consider two cases:
   - Case 1: \( x \le \frac{1}{\sqrt{2}} \)
     \[
     f(x, y) = \min \left( x, \frac{y}{x^2 + y^2} \right) \le x \le \frac{1}{\sqrt{2}}
     \]
   - Case 2: \( x \ge \frac{1}{\sqrt{2}} \)
     \[
     \text{We need to show that } \frac{y}{x^2 + y^2} \le \frac{1}{\sqrt{2}}
     \]
     Consider the inequality:
     \[
     \frac{y}{x^2 + y^2} \le \frac{1}{\sqrt{2}}
     \]
     This is equivalent to:
     \[
     y \le \frac{x^2 + y^2}{\sqrt{2}}
     \]
     Multiplying both sides by \(\sqrt{2}\):
     \[
     \sqrt{2} y \le x^2 + y^2
     \]
     Rearranging terms:
     \[
     y^2 - \sqrt{2} y + x^2 \ge 0
     \]
     This is a quadratic in \( y \). The discriminant of this quadratic is:
     \[
     (\sqrt{2})^2 - 4 \cdot 1 \cdot x^2 = 2 - 4x^2
     \]
     Since \( x \ge \frac{1}{\sqrt{2}} \), we have:
     \[
     2 - 4x^2 \le 0 \implies y^2 - \sqrt{2} y + x^2 \ge 0
     \]
     Therefore:
     \[
     \frac{y}{x^2 + y^2} \le \frac{1}{\sqrt{2}}
     \]
     Hence:
     \[
     f(x, y) = \min \left( x, \frac{y}{x^2 + y^2} \right) \le \frac{y}{x^2 + y^2} \le \frac{1}{\sqrt{2}}
     \]

5. Combining both cases, we have shown that:
   \[
   f(x, y) \le \frac{1}{\sqrt{2}} \text{ for all positive } x \text{ and } y
   \]

The final answer is \( \boxed{ \frac{1}{\sqrt{2}} } \).