In the triangle $ABC$,  $h_a, h_b, h_c$ are the altitudes   and $p$ is its half-perimeter. Compare $p^2$ with $h_ah_b + h_bh_c + h_ch_a$.

(Gregory Filippovsky)
1. **Express the altitudes in terms of the sides and the circumradius:**
   The altitude \( h_a \) from vertex \( A \) to side \( BC \) can be expressed as:
   \[
   h_a = \frac{2 \cdot \text{Area of } \triangle ABC}{a} = \frac{2 \cdot \frac{abc}{4R}}{a} = \frac{bc}{2R}
   \]
   Similarly, we have:
   \[
   h_b = \frac{ac}{2R} \quad \text{and} \quad h_c = \frac{ab}{2R}
   \]

2. **Calculate the sum \( h_bh_c + h_ch_a + h_ah_b \):**
   \[
   h_bh_c = \left( \frac{ac}{2R} \right) \left( \frac{ab}{2R} \right) = \frac{a^2bc}{4R^2}
   \]
   \[
   h_ch_a = \left( \frac{ab}{2R} \right) \left( \frac{bc}{2R} \right) = \frac{ab^2c}{4R^2}
   \]
   \[
   h_ah_b = \left( \frac{bc}{2R} \right) \left( \frac{ac}{2R} \right) = \frac{abc^2}{4R^2}
   \]
   Summing these up:
   \[
   h_bh_c + h_ch_a + h_ah_b = \frac{a^2bc}{4R^2} + \frac{ab^2c}{4R^2} + \frac{abc^2}{4R^2} = \frac{abc(a + b + c)}{4R^2}
   \]

3. **Express the sum in terms of the semi-perimeter \( p \):**
   The semi-perimeter \( p \) is given by:
   \[
   p = \frac{a + b + c}{2}
   \]
   Therefore:
   \[
   h_bh_c + h_ch_a + h_ah_b = \frac{abc \cdot 2p}{4R^2} = \frac{2p \cdot abc}{4R^2} = \frac{p \cdot abc}{2R^2}
   \]

4. **Relate the product \( abc \) to the inradius \( r \) and circumradius \( R \):**
   Using the formula for the area of the triangle \( \triangle ABC \):
   \[
   \text{Area} = \frac{abc}{4R} = pr
   \]
   Thus:
   \[
   abc = 4Rpr
   \]

5. **Substitute \( abc \) back into the sum:**
   \[
   h_bh_c + h_ch_a + h_ah_b = \frac{p \cdot 4Rpr}{2R^2} = \frac{4p^2r}{R}
   \]

6. **Compare \( p^2 \) with \( h_bh_c + h_ch_a + h_ah_b \):**
   According to Euler's inequality, \( R \geq 2r \). Therefore:
   \[
   \frac{4p^2r}{R} \leq \frac{4p^2r}{2r} = 2p^2
   \]
   Since \( 2r \leq R \), we have:
   \[
   \frac{4p^2r}{R} \leq p^2
   \]

Thus, we conclude:
\[
\boxed{h_bh_c + h_ch_a + h_ah_b \leq p^2}
\]