3. Find all values of $c$ for which the inequality $a+\sqrt{b+c}>b+\sqrt{a+c}$ holds for any positive $a, b$ and $a>b$.

(25 points.)
Answer: $c=\frac{1}{4}$.

Write the inequality in the form $a-b>\sqrt{a+c}-\sqrt{b+c}$. By multiplying and dividing the right part by its conjugate, we can represent the inequality as

$$
a-b>\frac{a-b}{\sqrt{a+c}+\sqrt{b+c}} \quad \Longleftrightarrow \quad \sqrt{a+c}+\sqrt{b+c}>1
$$

The last inequality must hold for any $a>b>0$. As the values of $a$ and $b$ increase, the left side only increases, so it is sufficient for the inequality to hold only for the smallest value of the expression $\sqrt{a+c}+\sqrt{b+c}$, which is obviously greater than $\sqrt{0+c}+\sqrt{0+c}=2 \sqrt{c}$. Therefore, $2 \sqrt{c}=1$, which means $c=\frac{1}{4}$.

Criteria. Only the answer -0 points. Proven that when $c=\frac{1}{4}$, the condition of the problem is satisfied -12 points.