If $a, b, c$ are positive real numbers, then the least value of $(a+b+c)(ab+bc+ca)$ is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $9abc$
Consider $(a+b+c)(ab+bc+ca)$ with $a,b,c>0$.
By AM–GM inequality,
$a+b+c \geq 3(abc)^{\frac{1}{3}} \quad ...(1)$
Similarly,
$ab+bc+ca \geq 3(abc)^{\frac{2}{3}} \quad ...(2)$
Multiplying (1) and (2),
$(a+b+c)(ab+bc+ca) \geq 9abc$
Equality holds when $a=b=c$.
therefore The least value of $(a+b+c)(ab+bc+ca)$ is $9abc$.