Question

Let side lengths of a triangle be `a`, `b`, and `c`. Then how do you proof that `a^2<2(b^2+c^2)`?

Answer 1

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)`

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