Problem 1432

1. Given: An acute triangle with unit area. We need to show that there is a point inside the triangle whose distance from each of the vertices is at least $\frac{2}{\sqrt[4]{27}}$.

2. Step 1: Consider the triangle with vertices $A, B,$ and $C$. Let $P$ be a point inside the triangle such that the distances from $P$ to $A, B,$ and $C$ are $PA, PB,$ and $PC$ respectively.

3. Step 2: We need to show that there exists a point $P$ such that $PA, PB,$ and $PC$ are all at least $\frac{2}{\sqrt[4]{27}}$.

4. Step 3: Use the fact that the area of the triangle is given by:
$$\text{Area} = \frac{1}{2} \times AB \times h$$
where $h$ is the height from vertex $C$ to side $AB$. Given that the area is 1, we have:
$$1 = \frac{1}{2} \times AB \times h \implies AB \times h = 2$$

5. Step 4: Consider the incenter $I$ of the triangle, which is the point where the angle bisectors of the triangle intersect. The incenter is equidistant from all sides of the triangle. Let $r$ be the inradius of the triangle.

6. Step 5: The area of the triangle can also be expressed in terms of the inradius $r$ and the semiperimeter $s$:
$$\text{Area} = r \times s$$
Given that the area is 1, we have:
$$r \times s = 1$$

7. Step 6: For an acute triangle, the inradius $r$ is less than the circumradius $R$. Using the fact that the area of the triangle is 1, we can relate the circumradius $R$ to the sides of the triangle using the formula:
$$R = \frac{abc}{4K}$$
where $K$ is the area of the triangle, and $a, b,$ and $c$ are the sides of the triangle. Given $K = 1$, we have:
$$R = \frac{abc}{4}$$

8. Step 7: Using the fact that $\sin A \sin B \sin C \leq \left(\frac{\sqrt{3}}{2}\right)^3$, we can show that:
$$\sqrt[3]{\sin A \sin B \sin C} \leq \frac{\sin A + \sin B + \sin C}{3} \leq \sin\left(\frac{A + B + C}{3}\right) = \frac{\sqrt{3}}{2}$$
This follows from Jensen's inequality for the concave function $\sin x$.

9. Step 8: Since the area of the triangle is 1, and using the properties of the incenter and the distances from the incenter to the vertices, we can conclude that there exists a point $P$ inside the triangle such that the distances from $P$ to each of the vertices are at least $\frac{2}{\sqrt[4]{27}}$.

10. Conclusion: Therefore, we have shown that there is a point inside the triangle whose distance from each of the vertices is at least $\frac{2}{\sqrt[4]{27}}$.

$\blacksquare$