Question

If ${\log }_2(a+b)+{\log }_2(c+d)\ge 4$, then minimum value of $a+b+c+d$ is

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

The correct option is A $8$

${\log }_2(a+b)+{\log }_2(c+d)\ge 4$
$\Rightarrow (a+b)(c+d)\ge 2^4$
As we know that,

$AM\ge GM$

Let assume numbers be $(a+b)$ and $(c+d)$, then

Using the property, we get
$\frac{a+b+c+d}{2}\ge \sqrt{(a+b)(c+d)}\ge 2^2$
$\Rightarrow a+b+c+d\ge 8$

Hence, option A.