Let $a$, $b$ and $m$ be three positive real numbers  and $a>b$. Which of the numbers $A=\sqrt{a+m}-\sqrt{a}$ and $B=\sqrt{b+m}-\sqrt{b}$ is bigger: 
1. Given that \(a, b, m\) are positive real numbers and \(a > b\), we need to compare \(A = \sqrt{a+m} - \sqrt{a}\) and \(B = \sqrt{b+m} - \sqrt{b}\).

2. Consider the expressions \(A\) and \(B\):
   \[
   A = \sqrt{a+m} - \sqrt{a}
   \]
   \[
   B = \sqrt{b+m} - \sqrt{b}
   \]

3. To compare \(A\) and \(B\), we can use the fact that the function \(f(x) = \sqrt{x}\) is concave. For a concave function, the difference \(\sqrt{x+m} - \sqrt{x}\) decreases as \(x\) increases.

4. Since \(a > b\), we have:
   \[
   \sqrt{a+m} + \sqrt{a} > \sqrt{b+m} + \sqrt{b}
   \]

5. Taking the reciprocals of both sides (since all terms are positive), we get:
   \[
   \frac{1}{\sqrt{b+m} + \sqrt{b}} > \frac{1}{\sqrt{a+m} + \sqrt{a}}
   \]

6. Multiplying both sides by \(m\) and using the difference of squares, we obtain:
   \[
   \frac{\sqrt{b+m} - \sqrt{b}}{m} > \frac{\sqrt{a+m} - \sqrt{a}}{m}
   \]

7. Simplifying, we get:
   \[
   \sqrt{b+m} - \sqrt{b} > \sqrt{a+m} - \sqrt{a}
   \]

8. Therefore, we conclude that:
   \[
   B > A
   \]

The final answer is \( \boxed{ B > A } \)