Question

If $a,b,c$ are distinct positive real numbers such that $a^2+b^2+c^2=1$, then value of $ab+bc+ca$, is:

• A. less than $1$
• B. $1$
• C. greater than $1$
• D. any real number

Answer 1

The correct option is A less than $1$

We know $2(ab+ac+bc)=(a+b+c{)}^2-a^2-b^2-c^2$
$2(ab+ac+bc)=(a+b+c{)}^2-1$
$(ab+bc+ac)=\frac{(a+b+c{)}^2-1}{2}$
Now if $a=b=c=\frac{1}{\sqrt{3}}$, then the above expression attains maximum value of 1.
Hence for distinct values of $a,b$ and $c$,
$\frac{(a+b+c{)}^2-1}{2}<1$
Hence, $ab+bc+ac<1$