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

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 $b , c \in \mathbb{R}$, ${b}^{2} + {c}^{2} \ge 2 b c$

As a short justification for (2):
${\left(b - c\right)}^{2} \ge 0$
$\implies {b}^{2} - 2 b c + {c}^{2} \ge 0$
$\implies {b}^{2} + {c}^{2} \ge 2 b c$

Claim: For a triangle with side lengths $a , b , c$, it is the case that ${a}^{2} < 2 \left({b}^{2} + {c}^{2}\right)$

Proof of Claim:

By (1), $a < b + c$

As $a > 0$ this implies
${a}^{2} < {\left(b + c\right)}^{2} = {b}^{2} + 2 b c + {c}^{2}$.

But by (2), $2 b c \le {b}^{2} + {c}^{2}$. Thus

${a}^{2} < {b}^{2} + \left(2 b c\right) + {c}^{2} \le {b}^{2} + \left({b}^{2} + {c}^{2}\right) + {c}^{2} = 2 \left({b}^{2} + {c}^{2}\right)$