Question

If $a,b\in R$, $a,b>0$, then $ab\le \frac{1}{4}{(a+b)}^2$,$a+b=1$

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)

$9$

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

$\Rightarrow 2\sqrt{ab}\le a+b$

$\Rightarrow (\sqrt{a}{)}^2+(\sqrt{b}{)}^2-2\sqrt{a}\sqrt{b}\ge 0$

$\Rightarrow (\sqrt{a}-\sqrt{b}{)}^2\ge 0$

$\Rightarrow \sqrt{a}\ge \sqrt{b}\Rightarrow a\ge b$

for least value $a=b$

$a+b=1$ and $a=b$

$\Rightarrow a=b=\frac{1}{2}$

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

$(1+\frac{1}{a})(1+\frac{1}{b})=\left(3\right)\times \left(3\right)$

$\therefore (1+\frac{1}{a})(1+\frac{1}{b})=9$