Question

Answer this question, based on the information given below.
If $a,bϵR,a,b>0$, then $ab\le \frac{1}{4}(a+b{)}^2$
If $a,b>0,$ $a+b=1$, then find the least value of $(1+\frac{1}{a})(1+\frac{1}{b})$

• A. $3$
• B. $6$
• C. $9$
• D. $12$

Answer 1

Answer: (C)

$9$

$(1+\frac{1}{a})(1+\frac{1}{b})$

$=\frac{(a+1)(b+1)}{ab}$

$=\frac{ab+a+b+1}{ab}$

$=1+\frac{a+b+1}{ab}$

$=1+\frac{2}{ab}$ ..... $(\because a+b=1)$

Now $ab\le \frac{1}{4}(a+b){}^2$

Therefore

$ab\le \frac{1}{4}\left(1\right){}^2$

$ab\le \frac{1}{4}$ ...(ii)

Hence, the maximum value of $ab=\frac{1}{4}$

Substituting in $\left(i\right)$, we get

$(1+\frac{1}{a})(1+\frac{1}{b})$

$=1+\frac{2}{ab}$

$\ge 1+\frac{2}{1/4}$ $=1+8$ $=9$

Hence, the answer is C.