Question

If $a^2+b^2+c^2=1$ and $p=ab+bc+ca$, then

• A. $\frac{1}{2}\le p<2$
• B. $-\frac{1}{2}\le p\le \frac{1}{2}$
• C. $-\frac{1}{2}\le p\le 1$
• D. $-1\le p\le \frac{1}{2}$

Answer 1

The correct option is C $-\frac{1}{2}\le p\le 1$

Given, $a^2+b^2+c^2=1$

We know that, $(a+b+c{)}^2\ge 0$

$=>a^2+b^2+c^2+2(ab+bc+ca)\ge 0$

$=>1+2(ab+bc+ca)\ge 0$

$=>ab+bc+ca\ge \frac{-1}{2}$

Again,

$(b-c{)}^2+(c-a{)}^2+(a-b{)}^2\ge 0$

$=>2(a^2+b^2+c^2)-2(ab+bc+ca)\ge 0$

$=>ab+bc+ca\le a^2+b^2+c^2$

$=>ab+bc+ca\le 1$

$=>p\le 1$ ...[Since $p=ab+bc+ca$]

Hence, $\frac{-1}{2}\le p\le 1$