Let $a$ and $b$ be two fixed positive real numbers. Find all real numbers $x$, such that inequality holds $$\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{a+b-x}} < \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}}$$
1. Define the function \( f(x) = \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{a+b-x}} \) for \( x \in (0, a+b) \).
2. To analyze the behavior of \( f(x) \), we first find its derivative:
   \[
   f'(x) = -\frac{1}{2} x^{-\frac{3}{2}} + \frac{1}{2} (a+b-x)^{-\frac{3}{2}}
   \]
3. Set the derivative to zero to find critical points:
   \[
   f'(x) = 0 \implies -\frac{1}{2} x^{-\frac{3}{2}} + \frac{1}{2} (a+b-x)^{-\frac{3}{2}} = 0
   \]
   \[
   \implies x^{-\frac{3}{2}} = (a+b-x)^{-\frac{3}{2}}
   \]
   \[
   \implies x = a+b-x
   \]
   \[
   \implies 2x = a+b
   \]
   \[
   \implies x = \frac{a+b}{2}
   \]
4. The critical point \( x = \frac{a+b}{2} \) divides the interval \( (0, a+b) \) into two parts: \( \left(0, \frac{a+b}{2}\right] \) and \( \left[\frac{a+b}{2}, a+b\right) \).
5. To determine the behavior of \( f(x) \) in these intervals, we examine the second derivative or the sign of the first derivative in each interval:
   - For \( x \in \left(0, \frac{a+b}{2}\right) \), \( f'(x) < 0 \), so \( f(x) \) is decreasing.
   - For \( x \in \left(\frac{a+b}{2}, a+b\right) \), \( f'(x) > 0 \), so \( f(x) \) is increasing.
6. Therefore, \( f(x) \) has a minimum at \( x = \frac{a+b}{2} \).
7. We need to find \( x \) such that \( f(x) < f(a) \) and \( f(x) < f(b) \):
   - Since \( f(x) \) is decreasing on \( \left(0, \frac{a+b}{2}\right] \) and increasing on \( \left[\frac{a+b}{2}, a+b\right) \), the values of \( f(x) \) at the endpoints \( a \) and \( b \) are higher than the values in between.
   - Thus, the inequality \( f(x) < f(a) \) and \( f(x) < f(b) \) holds for \( x \in (\min(a, b), \max(a, b)) \).

The final answer is \( \boxed{ x \in (\min(a, b), \max(a, b)) } \).