# Algebraic inequality

• Jun 16th 2008, 07:17 PM
mathwizard
Algebraic inequality
If $a, b, c$ are real numbers and $a + b + c = 0$, prove that $-1/2 \le ab + bc + ac \le 1$.

EDIT: Damned typo! The upper bound should be $1$ instead of $-1$!
• Jun 16th 2008, 07:31 PM
Isomorphism
Quote:

Originally Posted by mathwizard
If $a, b, c$ are real numbers and $a + b + c = 0$, prove that $-1/2 \le ab + bc + ac \le -1$.

$a + b + c = 0 \Rightarrow a^2 + b^2 + c^2 = -2(ab+bc+ca)$

WE have to prove $-1/2 \le ab + bc + ac \le -1 \Leftrightarrow 1 \ge -2(ab + bc + ac) \ge 2 \Leftrightarrow 1 \ge a^2 + b^2 + c^2 \ge 2$

But this is false, because we choose a= 1.03, b = -0.47, c = -0.56, a+b+c = 0, but $a^2 + b^2 + c^2 = 1.594 < 2$
• Jun 17th 2008, 01:28 AM
Moo
Hello,

Quote:

Originally Posted by Isomorphism
$a + b + c = 0 \Rightarrow a^2 + b^2 + c^2 = -2(ab+bc+ca)$

WE have to prove $-1/2 \le ab + bc + ac \le -1 \Leftrightarrow 1 \ge -2(ab + bc + ac) \ge 2 \Leftrightarrow 1 \ge a^2 + b^2 + c^2 \ge 2$

But this is false, because we choose a= 1.03, b = -0.47, c = -0.56, a+b+c = 0, but $a^2 + b^2 + c^2 = 1.594 < 2$

This is false because you can't have a number which is both superior to 2 and inferior to 1 (Rofl)

Like in #1, you can't have something that is both greater than -1/2 and smaller than -1.