Question

If $a^2+b^2+c^2=1$, then ab + bc + ca lies in the interval

• A. (1, 2)
• B. $[0,\frac{1}{2})$
• C. $[-\frac{1}{2},1]$
• D. [0, 1]

Answer 1

The correct option is C $[-\frac{1}{2},1]$

$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\sum (a-b{)}^2\ge 0$
or $1-\sum ab\ge 0\therefore \sum ab\le 1\dots \dots \left(1\right)$
$(a+b+c{)}^2=\sum a^2+2\sum ab=1+2\sum ab\ge 0$
$\therefore \sum ab\ge -\frac{1}{2}\dots \dots \left(2\right)$
Hence from (1) and (2), $(\sum ab)$ lies in $[-\frac{1}{2},1]$.