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

Question

If a, b, c, d are positive real numbers such that a + b +c + d = 2, then M =(a + b) (c+d) satisfies the relation. (A)  0 < M le 1 (B)  1 le M le 2 (C)  2 le M le 3  (D)  3 le M le 4

Tardigrade Admin · Accepted Answer

Since, AM ge GM, then ((a + b) + (c + d)/2) ge √ (a+ b) (c+d) ⇒ M le 1 Also, (a + b) + (c +d) > 0 [∵ a, b, c, d > 0] ∴ 0 < M le 1

Q. If $a, b, c, d$ are positive real numbers such that $a + b + c + d = 2,$ then $M = (a + b) (c + d)$ satisfies the relation.

$0 < M \leq 1$

$1 \leq M \leq 2$

$2 \leq M \leq 3$

$3 \leq M \leq 4$

Since, $A M \geq GM$ , then
$\frac{( a + b ) + ( c + d )}{2} \geq (a + b) (c + d) \Rightarrow M \leq 1$
Also, $(a + b) + (c + d) > 0 [∵ a, b, c, d > 0]$
$∴ 0 < M \leq 1$

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

0<M≤1

1≤M≤2

2≤M≤3

3≤M≤4