How to show that if $a^2+b^2+c^2=1$ then $\frac{-1}{2}\le ab+ac+bc \le 1$

Question

How to show that if $a^2+b^2+c^2=1$ then:

$$\frac{-1}{2}\le ab+ac+bc \le 1$$

From the assumption I see that:

$$-\frac{1}{2}\le\frac{\left(a+b+c\right)^{2}-1}{2}=\frac{\left(a+b+c\right)^{2}-\left(a^{2}+b^{2}+c^{2}\right)}{2}=ab+ac+bc $$ On the other hand $a^2=1-(b^2+c^2)<1$,then same does hold for $b^2$ and $c^2$ which shows that $-1<a<1$ and so :$$-3<a+b+c<3$$ $$\implies ab+ac+bc =\frac{\left(a+b+c\right)^{2}-1}{2}<4$$

But it's does not what I've been asked to prove,so how can I show that?

Do you know Lagrange multipliers ? – Claude Leibovici Dec 14 '20 at 15:06 — Claude Leibovici, Dec 14 '20 at 15:06
The second part is Cauchy-Schwarz inequality. – trancelocation Dec 14 '20 at 15:09 — trancelocation, Dec 14 '20 at 15:09

score 2 · Accepted Answer · answered Dec 14 '20 at 15:10

2

Hint: For the 2 inequalities you want to prove, expand the left hand sides of: \begin{align} (a+b+c)^2&\geq 0,\\ (a-b)^2+(b-c)^2+(c-a)^2&\geq 0. \end{align} And remember to use $a^2+b^2+c^2=1$.

answered Dec 14 '20 at 15:10

yurnero

10,505

How to show that if $a^2+b^2+c^2=1$ then $\frac{-1}{2}\le ab+ac+bc \le 1$

1 Answers1