If a, b and c are positive real numbers, then Match List-I with List-…

The correct answer is Option (2) → (A)-(I), (B)-(III), (C)-(II), (D)-(IV)

List-I (Expression)	List-II (The Least value of the expression)
(A) $(a + b)(b+c)(c+a)$	(I) $8abc$
(B) $(a+b+c)(ab+be+ca)$	(III) $9abc$
(C) $(a^2b+ b^2c + c^2a) (ab^2 + bc^2+ ca^2)$	(II) $9a^2b^2c^2$
(D) $(a + b)^2(b+c)^2 (c+a)^2$	(IV) $64a^2b^2c^2$

Since $a,b,c>0$, by AM–GM inequality the least values occur at $a=b=c$.

(A) $(a+b)(b+c)(c+a)$ at $a=b=c$ gives

$(2a)(2a)(2a)=8a^{3}=8abc$.

So (A) → (I).

(B) $(a+b+c)(ab+bc+ca)$ at $a=b=c$ gives

$(3a)(3a^{2})=9a^{3}=9abc$.

So (B) → (III).

(C) $(a^{2}b+b^{2}c+c^{2}a)(ab^{2}+bc^{2}+ca^{2})$ at $a=b=c$ gives

$(3a^{3})(3a^{3})=9a^{6}=9a^{2}b^{2}c^{2}$.

So (C) → (II).

(D) $(a+b)^{2}(b+c)^{2}(c+a)^{2}$ at $a=b=c$ gives

$(2a)^{2}(2a)^{2}(2a)^{2}=64a^{6}=64a^{2}b^{2}c^{2}$.

So (D) → (IV).

final answer: (A)–(I), (B)–(III), (C)–(II), (D)–(IV)