Let side lengths of a triangle be aa, bb, and cc. Then how do you proof that a^2<2(b^2+c^2)a2<2(b2+c2)?

1 Answer
Dec 19, 2015

We will use two facts:

  1. The sum of lengths of two sides of a triangle is greater than the length of the third side (this is known as the triangle inequality).

  2. For b, c in RR, b^2 + c^2 >= 2bc

As a short justification for (2):
(b-c)^2 >= 0
=> b^2 -2bc + c^2 >= 0
=> b^2 + c^2 >= 2bc

Claim: For a triangle with side lengths a, b, c, it is the case that a^2 <2(b^2 + c^2)

Proof of Claim:

By (1), a < b+c

As a > 0 this implies
a^2 < (b+c)^2 = b^2 + 2bc + c^2.

But by (2), 2bc <= b^2 + c^2. Thus

a^2 < b^2 + (2bc) + c^2 <= b^2 + (b^2 + c^2) + c^2 = 2(b^2 + c^2)