Question

if a, b, c are distinct +ve real numbers and $a^2+b^2+c^2=1$ then ab + bc + ca is

• A. less than $1$
• B. equal to $1$
• C. greater than $1$
• D. any real no.

Answer 1

The correct option is A less than $1$

In such type of problem if sum of the squares of number is known

and you need product of numbers taken two at a time or needed range of the product of numbers taken two at a time.

Start with square of the sum of the numbers like

$(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$

$2(ab+bc+ca)=(a+b+c{)}^2–(a^2+b^2+c^2)$

$(ab+bc+ca)=\frac{\left[\right(a+b+c{)}^2–1]}{2}$

Since $a^2+b^2+c^2=1,$ all $a,b$ and $c$ are less than $1$.

So, $(ab+bc+ca)$ must be less than $1.$