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 $\displaystyle -1$!

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- Jun 16th 2008, 07:17 PMmathwizardAlgebraic 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 PMIsomorphism
$\displaystyle a + b + c = 0 \Rightarrow a^2 + b^2 + c^2 = -2(ab+bc+ca)$

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 AMMoo