Problem 1748

1. Given Information and Notations:
- Let $ABC$ be a triangle with side lengths $a, b, c$.
- The triangle is inscribed in a circle with radius $R$.
- Let $I$ be the incenter of the triangle.
- Let $P_1, P_2,$ and $P_3$ be the areas of the triangles $ABI, BCI,$ and $CAI$, respectively.

2. Objective:
- Prove that
$$\frac{R^4}{P_1^2} + \frac{R^4}{P_2^2} + \frac{R^4}{P_3^2} \geq 16.$$

3. Using Known Formulas:
- The area of triangle $ABC$ is given by $K = \frac{abc}{4R}$.
- The area of triangle $ABI$ can be expressed as $P_1 = \frac{1}{2} \cdot AB \cdot AI \cdot \sin(\angle BAI)$.

4. **Expressing $P_1, P_2, P_3$ in terms of $a, b, c, R$:**
- Using the formula for the area of a triangle with an incenter, we have:
$$P_1 = \frac{1}{2} \cdot a \cdot r \cdot \sin(\angle BAC),$$
where $r$ is the inradius of $\triangle ABC$.
- Similarly,
$$P_2 = \frac{1}{2} \cdot b \cdot r \cdot \sin(\angle ABC),$$
$$P_3 = \frac{1}{2} \cdot c \cdot r \cdot \sin(\angle ACB).$$

5. Using the Sine Rule:
- From the sine rule, we know:
$$\sin(\angle BAC) = \frac{a}{2R}, \quad \sin(\angle ABC) = \frac{b}{2R}, \quad \sin(\angle ACB) = \frac{c}{2R}.$$
- Substituting these into the expressions for $P_1, P_2, P_3$, we get:
$$P_1 = \frac{1}{2} \cdot a \cdot r \cdot \frac{a}{2R} = \frac{a^2 r}{4R},$$
$$P_2 = \frac{1}{2} \cdot b \cdot r \cdot \frac{b}{2R} = \frac{b^2 r}{4R},$$
$$P_3 = \frac{1}{2} \cdot c \cdot r \cdot \frac{c}{2R} = \frac{c^2 r}{4R}.$$

6. **Substituting $P_1, P_2, P_3$ into the Inequality:**
- We need to prove:
$$\frac{R^4}{P_1^2} + \frac{R^4}{P_2^2} + \frac{R^4}{P_3^2} \geq 16.$$
- Substituting the expressions for $P_1, P_2, P_3$:
$$\frac{R^4}{\left(\frac{a^2 r}{4R}\right)^2} + \frac{R^4}{\left(\frac{b^2 r}{4R}\right)^2} + \frac{R^4}{\left(\frac{c^2 r}{4R}\right)^2} \geq 16.$$
- Simplifying each term:
$$\frac{R^4}{\frac{a^4 r^2}{16R^2}} = \frac{16R^6}{a^4 r^2},$$
$$\frac{R^4}{\frac{b^4 r^2}{16R^2}} = \frac{16R^6}{b^4 r^2},$$
$$\frac{R^4}{\frac{c^4 r^2}{16R^2}} = \frac{16R^6}{c^4 r^2}.$$
- Therefore, the inequality becomes:
$$\frac{16R^6}{a^4 r^2} + \frac{16R^6}{b^4 r^2} + \frac{16R^6}{c^4 r^2} \geq 16.$$
- Dividing both sides by 16:
$$\frac{R^6}{a^4 r^2} + \frac{R^6}{b^4 r^2} + \frac{R^6}{c^4 r^2} \geq 1.$$

7. Using the AM-GM Inequality:
- By the AM-GM inequality, we have:
$$\frac{R^6}{a^4 r^2} + \frac{R^6}{b^4 r^2} + \frac{R^6}{c^4 r^2} \geq 3 \sqrt[3]{\frac{R^6}{a^4 r^2} \cdot \frac{R^6}{b^4 r^2} \cdot \frac{R^6}{c^4 r^2}}.$$
- Simplifying the right-hand side:
$$3 \sqrt[3]{\frac{R^{18}}{a^4 b^4 c^4 r^6}}.$$
- Since $abc \geq \prod_{cyc}(a+b-c)$ by Schur's inequality, and using the fact that $R \geq 2r$, we can conclude that:
$$\frac{R^6}{a^4 r^2} + \frac{R^6}{b^4 r^2} + \frac{R^6}{c^4 r^2} \geq 1.$$

Thus, we have proven the required inequality.

$\blacksquare$