Question

If $a,b>0$, $a+b=1$, then the least value of $(1+\frac{1}{a})(1+\frac{1}{b})$, is

• A. $3$
• B. $6$
• C. $9$
• D. $12$

Answer 1

Answer: (C)

$9$

Given, $a+b=1$

we know that, $A.M.\ge G.M.$

$⟹\frac{a+b}{2}\ge \sqrt{ab}$

$⟹\frac{1}{2}\ge \sqrt{ab}$

$⟹\sqrt{ab}\le \frac{1}{2}$

squaring on both sides

$⟹ab\le \frac{1}{4}$ --------------(1)

Similarly

$⟹\frac{1+a+1+b}{2}\ge \sqrt{(1+a)(1+b)}$

$⟹\frac{2+(a+b)}{2}\ge \sqrt{(1+a)(1+b)}$

$⟹\frac{2+1}{2}\ge \sqrt{(1+a)(1+b)}$

$⟹\frac{3}{2}\ge \sqrt{(1+a)(1+b)}$

squaring on both sides

$⟹\frac{1}{(1+a)(1+b)}\le \frac{4}{9}$ ---------------(2)

multiplying (1) and (2) we get

$⟹\frac{ab}{(1+a)(1+b)}\le \frac{1}{4}\ast \frac{4}{9}$

$⟹\frac{ab}{(1+a)(1+b)}\le \frac{1}{9}$

$⟹\frac{1}{(1+a)(1+b)}\le \frac{1}{9ab}$

$⟹\frac{(1+a)(1+b)}{ab}\ge 9$

$⟹\frac{(1+a)}{a}\ast \frac{(1+b)}{b}\ge 9$

$⟹(1+\frac{1}{a})(1+\frac{1}{b})\ge 9$

Therefore, the minimum value of $(1+\frac{1}{a})(1+\frac{1}{b})$ is $9$