Algebraic inequality

**mathwizard** · June 16th 2008, 07:17 PM

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 $-1$ !

**Isomorphism** · June 16th 2008, 07:31 PM

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 $a^2 + b^2 + c^2 = 1.594 < 2$

**Moo** · June 17th 2008, 01:28 AM

Hello,

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

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

Thread: Algebraic inequality

LinkBack

Thread Tools

Display

Algebraic inequality

Similar Math Help Forum Discussions

[SOLVED] Help with algebraic manipulation of an inequality

Crazy algebraic inequality

Algebraic inequality. I couldn't resist one more.

Algebraic Inequality

Algebraic Inequality

Search Tags