Let A1A2A3 and B1B2B3 be triangles. If

p = A1A2+A2A3+A3A1+B1B2+ B2B3+B3B1 and

q = A1B1+A1B2+A1B3+A2B1+A2B2+A2B3+A3B1+A3B2 + A3B3,

prove that 3p ≤ 4q.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Eugenio Villaseca, Ph.D.)

We know that in a triangle, according to the triangle inequality, the sum of the two sides is greater than the third side. And in the case of a degenerate triangle, the sum is equal to length of the third side. Therefore, we have the following inequalities

A1B1 + A2B1 ≥ A1A2

A1B2 + A2B2 ≥ A1A2

A1B3 + A2B3 ≥ A1A2

A2B1 + A3B1 ≥ A2A3

A2B2 + A3B2 ≥ A2A3

A2B3 + A3B3 ≥ A2A3

A1B1 + A3B1 ≥ A1A3

A1B2 + A3B2 ≥ A1A3

A1B3 + A3B3 ≥ A1A3

B1A1 + B2A1 ≥ B1B2

B1A2 + B2A2 ≥ B1B2

B1A3 + B2A3 ≥ B1B2

B2A1 + B3A1 ≥ B2B3

B2A2 + B3A2 ≥ B2B3

B2A3 + B3A3 ≥ B2B3

B1A1 + B3A1 ≥ B1B3

B1A2 + B3A2 ≥ B1B3

B1A3 + B3A3 ≥ B1B3

Now add all these inequalities, we have 3p ≤ 4q.