Question

lf $a,b,c$ are the sides of a triangle such that $a^2+b^2+c^2=1$ then the maximum value of $ab+bc+ca$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{5}{6}$

Answer 1

Answer: (A)

$1$

$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=2[(a^2+b^2+c^2)-(ab+bc+ac)]$
$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=2[1-(ab+bc+ac)]$
The L.H.S is the sum of three squares, hence it must be non-negative.
$\Rightarrow 2[1-(ab+bc+ac\left)\right]\ge 0$
$ab+bc+ac\le 1$
Hence, option A is the correct answer.