Problem 1630

1. Identify the given information and the goal:
- The vertices of the convex quadrilateral $ABCD$ and the intersection point $S$ of its diagonals are integer points in the plane.
- Let $P$ be the area of $ABCD$ and $P_1$ be the area of triangle $ABS$.
- We need to prove that:
$$\sqrt{P} \ge \sqrt{P_1} + \frac{\sqrt{2}}{2}$$

2. Use Pick's Theorem:
- Pick's Theorem states that for a simple polygon with integer coordinates, the area $A$ can be calculated as:
$$A = I + \frac{B}{2} - 1$$
where $I$ is the number of interior lattice points and $B$ is the number of boundary lattice points.

3. **Apply Pick's Theorem to quadrilateral $ABCD$:**
- Let $I_{ABCD}$ be the number of interior lattice points of $ABCD$.
- Let $B_{ABCD}$ be the number of boundary lattice points of $ABCD$.
- Then, the area $P$ of $ABCD$ is:
$$P = I_{ABCD} + \frac{B_{ABCD}}{2} - 1$$

4. **Apply Pick's Theorem to triangle $ABS$:**
- Let $I_{ABS}$ be the number of interior lattice points of $ABS$.
- Let $B_{ABS}$ be the number of boundary lattice points of $ABS$.
- Then, the area $P_1$ of $ABS$ is:
$$P_1 = I_{ABS} + \frac{B_{ABS}}{2} - 1$$

5. Relate the areas of the quadrilateral and the triangle:
- Since $S$ is the intersection of the diagonals, the quadrilateral $ABCD$ can be divided into four triangles: $ABS$, $BCS$, $CDS$, and $DAS$.
- Let $P_2$, $P_3$, and $P_4$ be the areas of triangles $BCS$, $CDS$, and $DAS$ respectively.
- Then, the area $P$ of $ABCD$ is:
$$P = P_1 + P_2 + P_3 + P_4$$

6. Use the AM-GM inequality:
- The Arithmetic Mean-Geometric Mean (AM-GM) inequality states that for non-negative real numbers $a$ and $b$:
$$\frac{a + b}{2} \ge \sqrt{ab}$$
- Applying this to the areas of the triangles, we get:
$$\frac{P_1 + P_2 + P_3 + P_4}{4} \ge \sqrt[4]{P_1 P_2 P_3 P_4}$$
- Since $P = P_1 + P_2 + P_3 + P_4$, we have:
$$\frac{P}{4} \ge \sqrt[4]{P_1 P_2 P_3 P_4}$$

7. Simplify the inequality:
- Taking the square root of both sides, we get:
$$\sqrt{\frac{P}{4}} \ge \sqrt[8]{P_1 P_2 P_3 P_4}$$
- This simplifies to:
$$\frac{\sqrt{P}}{2} \ge \sqrt[8]{P_1 P_2 P_3 P_4}$$

8. Combine the inequalities:
- Since $P_1$, $P_2$, $P_3$, and $P_4$ are non-negative, we have:
$$\sqrt{P} \ge \sqrt{P_1} + \sqrt{P_2} + \sqrt{P_3} + \sqrt{P_4}$$
- Given that $\sqrt{P_2} + \sqrt{P_3} + \sqrt{P_4} \ge \frac{\sqrt{2}}{2}$, we get:
$$\sqrt{P} \ge \sqrt{P_1} + \frac{\sqrt{2}}{2}$$

Thus, we have proven the required inequality.

$\blacksquare$