Question

If $a,b,c,d\in +R$ such that $a+b+c+d=2$, then $M=(a+b)(c+d)$ satisfies the relation

• A. $0\le M\le 1$
• B. $1\le M\le 2$
• C. $2\le M\le 3$
• D. $3\le M\le 4$

Answer 1

The correct option is A $0\le M\le 1$

Given, $a,b,c,d\in +R⟹a\ge 0,b\ge 0,c\ge 0,d\ge 0$

Therefore, $a+b\ge 0,c+d\ge 0$

Given $a+b+c+d=2$ and

$M=(a+b)(c+d)\ge 0$. Since, $a+b\ge 0,c+d\ge 0$

Therefore, $M\ge 0$ ————(1)

We know that $A.M.\ge G.M.$

$⟹\frac{(a+b)+(c+d)}{2}\ge \sqrt{(a+b)(c+d)}$

$⟹\frac{2}{2}\ge \sqrt{M}$

$⟹\sqrt{M}\le 1$

Squaring on both sides

$⟹M\le 1$ ————-(2)

From (1) and (2)

$0\le M\le 1$