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)

Each least value comes from the AM-GM inequality: for positive numbers, the arithmetic mean is at least the geometric mean, with equality when the numbers are equal (here a = b = c).

(A) \((a + b)(b+c)(c+a)\)

This matches (I) \(8abc\).

    \( \displaystyle a + b \ge 2\sqrt{ab},\quad b + c \ge 2\sqrt{bc},\quad c + a \ge 2\sqrt{ca}\)

    \( \displaystyle (a + b)(b + c)(c + a) \ge 8\sqrt{a^2b^2c^2} = 8abc\)

(B) \((a+b+c)(ab+bc+ca)\)

This matches (III) \(9abc\).

    \(a + b + c \ge 3(abc)^{1/3}\)
\(ab + bc + ca \ge 3(a^2b^2c^2)^{1/3}\)

    \((a + b + c)(ab + bc + ca) \ge 9(a^3b^3c^3)^{1/3} = 9abc\)

(C) \((a^2b+ b^2c + c^2a) (ab^2 + bc^2+ ca^2)\)

This matches (II) \(9a^2b^2c^2\). The product of the three terms in each bracket is \(a^3b^3c^3\):

    \(a^2b + b^2c + c^2a \ge 3(a^3b^3c^3)^{1/3} = 3abc\)
\(ab^2 + bc^2 + ca^2 \ge 3abc\)

    \((a^2b + b^2c + c^2a)(ab^2 + bc^2 + ca^2) \ge 9a^2b^2c^2\)

(D) \((a + b)^2(b+c)^2 (c+a)^2\)

This matches (IV) \(64a^2b^2c^2\). It is the square of the expression in (A):

    \((a + b)^2(b + c)^2(c + a)^2 \ge (8abc)^2 = 64a^2b^2c^2\)

So the matching is (A)-(I), (B)-(III), (C)-(II), (D)-(IV).

Shortcut: Put a = b = c = 1, where every AM-GM bound becomes an equality: (A) gives 8, (B) gives 9, (C) gives 9 and (D) gives 64. The List-II entries at a = b = c = 1 are 8, 9, 9 and 64, and the degrees separate (B) and (C): (B) has degree 3, so it matches \(9abc\); (C) has degree 6, so it matches \(9a^2b^2c^2\).