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

The minimum value of $(a+b+c)(ab+bc+ca)$ needs to be found when $a, b, c$ are positive real numbers.

Using AM-GM inequality on $(a+b+c)$:

$\frac{a+b+c}{3} \geq \sqrt[3]{abc}$

$a+b+c \geq 3\sqrt[3]{abc}$

Using AM-GM inequality on $(ab+bc+ca)$:

$\frac{ab+bc+ca}{3} \geq \sqrt[3]{ab \cdot bc \cdot ca}$

$\frac{ab+bc+ca}{3} \geq \sqrt[3]{a^2b^2c^2}$

$\frac{ab+bc+ca}{3} \geq (abc)^{2/3}$

$ab+bc+ca \geq 3(abc)^{2/3}$

Combining the two inequalities:

$(a+b+c)(ab+bc+ca) \geq 3\sqrt[3]{abc} \times 3(abc)^{2/3}$

$(a+b+c)(ab+bc+ca) \geq 9 \cdot (abc)^{1/3} \cdot (abc)^{2/3}$

$(a+b+c)(ab+bc+ca) \geq 9 \cdot (abc)^{1/3+2/3}$

$(a+b+c)(ab+bc+ca) \geq 9abc$

Equality holds in AM-GM when all terms are equal, so when $a = b = c$:

$(a+b+c) = 3a$

$(ab+bc+ca) = 3a^2$

$(a+b+c)(ab+bc+ca) = 3a \times 3a^2 = 9a^3 = 9abc$

Therefore, the least value of $(a+b+c)(ab+bc+ca)$ is $9abc$.