Question

If a, b, c, d are positive reals such that a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation

(a) 0 < M < 1 

(b) 1 < M < 2 

(c) 2 < M < 3 

(d) 3 < M < 4

Answer 1

(a) 0 < M < 1

Let x = a + b, y = c + d. Then, x + y = 1 and M = xy. 

If the sum of two quantities is a constant, then their product is maximum when both the quantities are equal, i.e., x = y 

⇒ x = 1, y = 1 ⇒ xy = 1. 

∴ Maximum value of M is 1 

Also, a, b, c, d being positive real numbers 

(a + b) (c + d) > 0 

∴ 0 < M < 1 

Alternatively: For positive quantities, AM > GM 

\(\frac12\){(a+b) + (c+d)} ≥ {(a+b) + (c+d)}^\(\frac12\)

⇒ \(M^{\frac12}\) ≤ \(\frac12\) x 2 ⇒ \(M^{\frac12}\) ≤ 1 ⇒ M < 1 

Also, M = (a + b) (c + d) > 0 

∴ 0 < M < 1.