Find the minimum.

Level 1

If $a$ and $b$ are rational positive numbers such that $a+b=1$ ,find the smallest possible value of $(1+\dfrac{1}{a})(1+\dfrac{1}{b})$ .

The answer is 9.

1 solution

Lorenc Bushi
Jan 3, 2014

Expanding the brackets and we get :

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

Since $a+b=1$ then it follows that $(1+\frac{1}{a})(1+\frac{1}{b})=1+\frac{2}{ab}$ ,Thus,we must find the minimum of this expression.In order to find the minimum of this expression we must first find the maximum of $ab$ .Since we know that $a+b=1$ the first thing that comes into my mind is $AM≥GM$ inequality.Let's take the specific numbers $a^2$ and $b^2$ and let's apply to them the $AM≥GM$ inequality.We should have:

$\frac{a^2+b^2}{2}≥\sqrt{a^2b^2}$ , and by speeding things up we get:

$a^2+b^2≥2ab$ .Rewrite the left side and we have:

$(a+b)^2-2ab≥2ab$ $\rightarrow$ $1-2ab≥2ab$ $\rightarrow$ $\boxed{ab≤\frac{1}{4}}$ , and by substituting this value for $ab$ we get the minimum of our expression which is $1+\frac{2}{\frac{1}{4}}=\boxed{9}$

I managed to make it right only in my third attempt. I wonder where my other answers went wrong .

Elaborately speaking , my $1st$ approach to the problem was :

Applying $A.M \geq G.M$ to the sets : $\left(\frac{1}{a},1\right),\left(\frac{1}{b},1\right)$ and $(a^2,b^2)$ separately we get,

$\frac{1}{a}+1\geq 2\sqrt{\frac{1}{a}}---(i)$

$\frac{1}{b}+1\geq 2\sqrt{\frac{1}{b}}---(ii)$

$a^2+b^2\geq 2ab---(iii)$

Now, $(i) \times (ii)$ gives,

$\left(\frac{1}{a}+1 \right)\left(\frac{1}{b}+1 \right) \geq 4\sqrt{\frac{1}{ab}}---(iv)$

To get the minimum value of the Left hand expression,clearly the $ab$ has to be maximised as much as possible.

From $(iii)$ and from the given info( $a+b=1$ ),it's easy to deduce the fact that :

$ab \leq \frac{1}{4},$ i.e the maximum value of $ab$ is $\frac{1}{4}$ .

Hence, the minimum value of the $\left(\frac{1}{a}+1 \right)\left(\frac{1}{b}+1 \right)$ occurs at $ab=\frac{1}{4}$ which is $\boxed{8}$ [Clearly achievable by putting value of $ab$ in the right hand expression of $(iv)].$

My $2nd$ approach was expanding the Left hand expression of $(iv)$ and getting the expression :

$\frac{2+ab}{ab} \geq 4\sqrt{\frac{1}{ab}}$

Since,our job is to find the minimum value of the expression, it's clearly equivalent to find when the equality exists or precisely for what value of $ab$ equality exists. So,squaring both sides,cancelling $ab$ from the denominators as clearly $ab \neq 0$ ,we get a quadratic equation where variable is $ab$ .Solving the equation yields $ab=6 \pm 4\sqrt{2}$ . Hence we need to take just $ab=6+4\sqrt{2}$ as $ab$ has to be more. So, putting this value of $ab$ in Right hand expression of $(iv)$ we get the minimum value of $\left(\frac{1}{a}+1 \right)\left(\frac{1}{b}+1 \right)$ as $\boxed{2(2-\sqrt{2})}$ . So,I would be glad if someone point out my mistakes if there are any.

Thanks.

Bhargav Das - 7 years, 5 months ago

In your first method, in (i) equality occurs at a = 1, in (ii) equality occurs at b = 1 ....since a=1, b=1 NOT possible simultaneosly (as a+b given as 1), your equation (iii) is not strict: i mean equality in (iii) is not achievable.

Such is not problem with solution by Lorenc.

To summarize: "whether Equality is achievable or not" needs always be checked in questions of finding Minimum or Maximum value using Inequalities.

Piyushkumar Palan - 7 years, 5 months ago

This was precisely my first attempt.... :p ....Cheers

Eddie The Head - 7 years, 4 months ago

Awesome question! Were you the one who made it? BTW, Nice solution.:)

Priyansh Sangule - 7 years, 5 months ago

Find the minimum.

The answer is 9.

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