How to prove that $4x \le x^2 + 8$ for all $x$

Question

I need to find a way to prove the above statement, and feel that doing it by individual cases is not the best method.

score 3 · Accepted Answer · answered Oct 17 '15 at 19:32

3

You may just write $$ x^2-4x+8=(x-2)^2+4\geq4>0 $$

answered Oct 17 '15 at 19:32

Olivier Oloa

score 1 · Answer 2 · answered Oct 17 '15 at 19:31

1

$x^2-4x+8$ has no real roots, so if it's positive somewhere, it must be positive everywhere. (As a consequence of intermediate value theorem.)

answered Oct 17 '15 at 19:31

T.J. Gaffney

2 Answers2