1. ## 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 $\displaystyle -1$!

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

3. Hello,

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

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