In the tetrahedron $ABCD,\angle BDC=90^o$ and the foot of the perpendicular from $D$ to $ABC$ is the intersection of the altitudes of $ABC$. Prove that: \[ (AB+BC+CA)^2\le6(AD^2+BD^2+CD^2). \] When do we have equality?
1. **Identify the orthocenter and relevant distances:**
   Let \( H \) be the orthocenter of \( \triangle ABC \). By definition, \( H \) is the intersection of the altitudes of \( \triangle ABC \). Given that \( \angle BDC = 90^\circ \), we can use the Pythagorean theorem in various triangles formed by \( D \).

2. **Use the Pythagorean theorem in triangles involving \( D \):**
   \[
   \begin{aligned}
   &AH^2 + HD^2 = AD^2 \quad \text{(1)} \\
   &BH^2 + HD^2 = BD^2 \quad \text{(2)} \\
   &CH^2 + HD^2 = CD^2 \quad \text{(3)} \\
   &BD^2 + CD^2 = BC^2 \quad \text{(4)}
   \end{aligned}
   \]

3. **Combine equations to eliminate \( HD \):**
   Adding equations (2) and (3):
   \[
   BH^2 + HD^2 + CH^2 + HD^2 = BD^2 + CD^2
   \]
   Using equation (4):
   \[
   BH^2 + CH^2 + 2HD^2 = BC^2
   \]
   Therefore:
   \[
   2HD^2 = BC^2 - BH^2 - CH^2
   \]

4. **Relate \( AD^2 \) to other sides:**
   Adding equations (1) and (2):
   \[
   AH^2 + HD^2 + BH^2 + HD^2 = AD^2 + BD^2
   \]
   Using the previous result:
   \[
   AH^2 + BH^2 + 2HD^2 = AD^2 + BD^2
   \]
   Substituting \( 2HD^2 \):
   \[
   AH^2 + BH^2 + (BC^2 - BH^2 - CH^2) = AD^2 + BD^2
   \]
   Simplifying:
   \[
   AH^2 + BC^2 - CH^2 = AD^2 + BD^2
   \]
   Since \( AH^2 + BC^2 - CH^2 = AB^2 \):
   \[
   AD^2 + BD^2 = AB^2
   \]

5. **Sum the squared distances:**
   Similarly, we can show:
   \[
   CD^2 + AD^2 = CA^2
   \]
   Adding these results:
   \[
   AD^2 + BD^2 + CD^2 = \frac{1}{3}(AB^2 + BC^2 + CA^2)
   \]

6. **Multiply by 3 and compare:**
   \[
   6(AD^2 + BD^2 + CD^2) = 2(AB^2 + BC^2 + CA^2)
   \]
   By the Cauchy-Schwarz inequality:
   \[
   (AB + BC + CA)^2 \leq 3(AB^2 + BC^2 + CA^2)
   \]
   Therefore:
   \[
   (AB + BC + CA)^2 \leq 6(AD^2 + BD^2 + CD^2)
   \]

7. **Equality condition:**
   Equality holds when \( \triangle ABC \) is equilateral, as this ensures the maximum symmetry and equal distribution of distances.

\(\blacksquare\)

The final answer is \( \boxed{ (AB + BC + CA)^2 \leq 6(AD^2 + BD^2 + CD^2) } \)