Problem 5

In tetrahedron

ABCD

\angle BDC=90^\circ

. The foot

H

of the perpendicular from

D

to plane

ABC

is the intersection of the altitudes of

\triangle ABC

. Prove that

(AB+BC+CA)^2\le6(AD^2+BD^2+CD^2)

. For what tetrahedra does equality hold?

Step 4 of 4: Apply Cauchy–Schwarz and identify equality

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(AB+BC+CA)^2\le3(AB^2+BC^2+CA^2)=6(AD^2+BD^2+CD^2)

Detailed analysis

Cauchy–Schwarz applied to $(AB,BC,CA)$ and $(1,1,1)$ gives $(AB+BC+CA)^2\le3(AB^2+BC^2+CA^2)$ . Using the identity above proves the required inequality. Equality holds exactly when $AB=BC=CA$ , so the base triangle ABC is equilateral.