Find the smallest real constant $p$ for which the inequality holds $\sqrt{ab}- \frac{2ab}{a + b} \le p \left( \frac{a + b}{2} -\sqrt{ab}\right)$ with any positive real numbers $a, b$.
1. We start with the given inequality:
   \[
   \sqrt{ab} - \frac{2ab}{a + b} \le p \left( \frac{a + b}{2} - \sqrt{ab} \right)
   \]
   for any positive real numbers \(a\) and \(b\).

2. Substitute \(a = 1\) and \(b = c^2\) where \(c \neq 1\):
   \[
   \sqrt{1 \cdot c^2} - \frac{2 \cdot 1 \cdot c^2}{1 + c^2} \le p \left( \frac{1 + c^2}{2} - \sqrt{1 \cdot c^2} \right)
   \]
   Simplifying, we get:
   \[
   c - \frac{2c^2}{1 + c^2} \le p \left( \frac{1 + c^2}{2} - c \right)
   \]

3. Simplify the left-hand side:
   \[
   c - \frac{2c^2}{1 + c^2} = \frac{c(1 + c^2) - 2c^2}{1 + c^2} = \frac{c + c^3 - 2c^2}{1 + c^2} = \frac{c(c^2 - 2c + 1)}{1 + c^2}
   \]

4. Simplify the right-hand side:
   \[
   p \left( \frac{1 + c^2}{2} - c \right) = p \left( \frac{1 + c^2 - 2c}{2} \right) = p \left( \frac{(c - 1)^2}{2} \right)
   \]

5. The inequality becomes:
   \[
   \frac{c(c^2 - 2c + 1)}{1 + c^2} \le p \left( \frac{(c - 1)^2}{2} \right)
   \]

6. To find the smallest \(p\), we need:
   \[
   p \ge \frac{2c}{c^2 + 1}
   \]

7. We want \(pc^2 - 2c + p \ge 0\) for all \(c \in \mathbb{R} - \{1, -1\}\). The discriminant of this quadratic must be non-negative:
   \[
   (2)^2 - 4p^2 \le 0 \implies 4 - 4p^2 \le 0 \implies p^2 \ge 1 \implies p \ge 1
   \]

8. We now prove the inequality for \(p = 1\):
   \[
   \sqrt{ab} - \frac{2ab}{a + b} \le \frac{a + b}{2} - \sqrt{ab}
   \]

9. Let \(a + b = x\) and \(ab = y^2\). The inequality becomes:
   \[
   \sqrt{y^2} - \frac{2y^2}{x} \le \frac{x}{2} - y
   \]
   Simplifying, we get:
   \[
   y - \frac{2y^2}{x} \le \frac{x}{2} - y
   \]

10. Rearrange the terms:
    \[
    2y \le \frac{x}{2} + \frac{2y^2}{x}
    \]

11. Multiply through by \(2x\):
    \[
    4xy \le x^2 + 4y^2
    \]

12. Rearrange to:
    \[
    x^2 - 4xy + 4y^2 \ge 0
    \]

13. This simplifies to:
    \[
    (x - 2y)^2 \ge 0
    \]
    which is always true.

\(\blacksquare\)

The final answer is \( \boxed{ p = 1 } \)