Question

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

• A. $[0,1]$
• B. $[-0.5,1]$
• C. $[0,0.5]$
• D. $[1,2]$

Answer 1

Answer: (B)

$[-0.5,1]$

we know that

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

$(a+b+c{)}^2\ge 0$ for any real a,b,c

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

Therefore, $1+2(ab+bc+ca)\ge 0$

$(ab+bc+ca)\ge -\frac{1}{2}$

Since, $A.M.\ge G.M.$

$⟹\frac{a+b}{2}\ge \sqrt{ab}$

$⟹a+b\ge 2\sqrt{ab}$

Assume $a=a^2$ and $b=b^2$

$⟹a^2+b^2\ge 2ab$ ------(1)

similarly,

$b^2+c^2\ge 2bc$ ------(2)

$c^2+a^2\ge 2ac$ ------(3)

adding (1), (2) and (3) we get

$a^2+b^2+c^2\ge ab+bc+ca$

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

$(ab+bc+ca)\le 1$

Therefore, $ab+bc+ca$ lies in the interval $[-\frac{1}{2},1]$