1. 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 $-1$!

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

3. 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.