Minimum value of of $a+b+\frac{1}{ab}$

Question

Okay so the question is:

If $a,b$ are positive real numbers satisfying $a^2+b^2=1$, then the minimum value of $a + b + \frac{1}{ab}$ is?

The answer (without any reasoning) provided by the book is $2+\sqrt 2$. So I assume the reason would be that, if we draw a circle in a $a-b$ 2d plane then since $a, b$ are positive we take the points $(\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$ and put in the equation so then we get that result.

But here is how I did it initially: Use AM-GM inequality then $a + b + \frac{1}{ab} \geq 3$

Then we can clearly see $3$ is smaller than $2+\sqrt 2$.

Am I missing something here?? Thanks in advance.

Answer 1

First of all, there is no contradiction between the two approaches... The minimum value of $2+\sqrt{2}$ is inded $\ge 3$. You can get the minimum by using the method of lagrange multipliers. The (local) minima will be stationary points of $$ F(a,b,\lambda) = a + b + \frac{1}{ab} - \lambda(a^2+b^2-1). $$

You can compute them and get the mentioned result.

Answer 2

Solution 1: as a function of $ab$:

Write $$f(a,b) = a+b+\frac{1}{ab} = \sqrt{(a+b)^2}+\frac{1}{ab}\\ = \sqrt{a^2 + b^2 + 2 ab}+\frac{1}{ab} = \sqrt{1 + 2 ab}+\frac{1}{ab} $$

since the restriction $a^2 + b^2 = 1$. So we see that $f$, under this restriction, only depends on the product $ab$.

The maximum value that $ab$ can attain under the restriction $a^2 + b^2 = 1$ is given by AM-GM, as $1 = a^2 + b^2 \ge 2 ab$, hence $ab \le \frac12$.

Now note that the derivative
$$\frac{d \; f(ab)}{d \; (ab)} = \frac{1}{\sqrt{1 + 2 ab}}- \frac{1}{(ab)^2} $$ is negative for all values $0 < ab \le \frac12$. Hence the minimum of $f(a,b)$ under the restriction $a^2 + b^2 = 1$ is attained at the maximum value for $ab$ which is $ab = \frac12$. Then we obtain

$$f(a,b) =\sqrt{1 + 2 ab}+\frac{1}{ab} = 2 + \sqrt{2}$$

$\qquad \Box$

Solution 2: with homogenization (with questions, see comments)

You can homogenize this to: $$f(a,b) = a+b+\frac{a^2+b^2}{ab}$$

Now you can demand some value for $b$, e.g. $b = 1$, which gives a multiple of $f(a,b)$ which satisfies the condition $a^2+b^2 = 1$ automatically, i.e.

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

By $F'(a) = 2 - \frac{1}{a^2} = 0$ you get at the minimum $a = \sqrt{1/2}$ immediately, which gives, for the original case, $b = \sqrt{1 - a^2} = \sqrt{1/2}$, and for the minimum function value you get $a+b+\frac{1}{ab} = 2 \sqrt{1/2} + 2 = \sqrt{2} + 2$. $\qquad \Box$