If $a, b, c$ are positive real numbers, then the least value of $(a+b+c)(ab+bc+ca)$ is:

$9abc$

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 $9abc$.