What is the minimum value of $a+b+\frac{1}{ab}$ if $a^2 + b^2 = 1$? For the case when $a,b>0,$ I used AM-GM Inequality as follows that:
$\frac{(a+b+\frac{1}{ab})}{3} \geq (ab\frac{1}{ab})^\frac{1}{3}$
This implies that $(a+b+\frac{1}{ab})\geq 3$. Hence, the minimum value of $(a+b+\frac{1}{ab})$ is 3
But the answer is $2+\sqrt{2}$ ... how is it ?

$2ab = (a+b)^2 - (a^2+b^2)$, hence we need to find the minimum value of $a + b + \frac{2}{(a+b)^2 - (a^2+b^2)}$ $ = a + b + \frac{2}{(a+b)^2 - 1}$.
Let us find the maximum value (k) of $a+b$ given $a^2+b^2=1$. Thinking graphically, when $a+b = k$ is tangent to the circle, its gradient is $-1$ and hence the gradient of the normal is $1$, or $a=b$. Hence $k = \frac{1}{\sqrt2} + \frac{1}{\sqrt2} = \sqrt{2}$, which gives $\sqrt{2} + \sqrt{2} + \frac{2}{2-1} = 2 + \sqrt{2}$ as the answer.
This relies on the fact that $f(x) = x + \frac{2}{x^2 - 1}$ is decreasing for $x \in (1, \sqrt{2}]$. Partial fractions gives $f(x) = x + \frac{1}{x-1} - \frac{1}{x+1}$ and taking the derivative shows that $f'(x) < 0$ in the given interval. There might be a way without calculus.
