# Algebraic inequality

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

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

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

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

WE have to prove \$\displaystyle -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 \$\displaystyle a^2 + b^2 + c^2 = 1.594 < 2\$
• Jun 17th 2008, 01:28 AM
Moo
Hello,

Quote:

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

WE have to prove \$\displaystyle -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 \$\displaystyle 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.