$xy=1 \implies $minimum $x+y=$?

Arithmetic Mean $\ge$ Geometric Mean

$\frac{x+y}{2} \ge \sqrt{xy} $

$ \implies x + y \ge 2 $

$ \implies $ Minimum Value = $2$

$x+y=x+y-2\sqrt{xy}+2\sqrt{xy}=(\sqrt{x}-\sqrt{y})^2+2\ge2$

The value will be attained when $(\sqrt{x}-\sqrt{y})^2=0\Rightarrow \sqrt{x}=\sqrt{y}\Rightarrow x=y=1$

Please note that this is the basis for the inequality of A.M.$\ge$ G.M.

If $x,y\ge 0$ then we know that $(\sqrt{x}-\sqrt{y})^2\ge0$

$\Rightarrow x+y-2\sqrt{xy}\ge 0$

$\Rightarrow x+y\ge 2\sqrt{xy}$

With $x\cdot y=1$ you know that $x=\frac{1}{y}$ so $$x+y=x+\frac{1}{x}$$ Wlog you can say now $x\geq 1 $, try to find the minimum now.

Another way is for x>0 we know that $$\frac{(x-1)^2 }{x}\geq 0$$ on the other hand $$\frac{(x-1)^2}{x} = \frac{x^2}{x} - \frac{2x}{x} +\frac{1}{x} =x-2 +\frac{1}{x} \geq 0$$ This is equivalent to $$x+\frac{1}{x}\geq 2$$ and as $1+1=2$ we know that 2 is the minimum.

I guess I get some downvotes for the following, but maybe it gives you some hope. If you don't like adding a smart zero (which is a bit unintuitive), you can still use advanced calculus methods. You try to minimize the function $f:\mathbb{R}^2\rightarrow \mathbb{R}; \ (x,y)\mapsto x+y$ with the side condition $x\cdot y = 1$.

Now you use a lagrangian multiplier and minimizing the function $$g(x,y)=x+y+\lambda (xy-1)$$ Now you take partial derivatives, which must be zero for a minimum \begin{align*} \frac{\partial g}{\partial x} &= 1+ \lambda y\\ \frac{\partial g}{\partial y} &= 1+ \lambda x\\ \frac{\partial g}{\partial \lambda} &= xy-1 \end{align*} Now you only need to solve the system \begin{align*} 0&=1+\lambda y\\ 0&= 1+\lambda x \\ 0&=xy - 1 \end{align*}

As Winston mentioned you could just take the derivative of $\displaystyle x+\frac{1}{x}$ which is $$1-\frac{1}{x^2}$$ So you need to solve the equation $$1-\frac{1}{x^2}=0 \iff 1=\frac{1}{x^2} \iff x^2=1$$

Note that $y=1/x$ and $$\left(x+\frac{1}{x}\right)^2=\left( x-\frac{1}{x}\right)^2+4\cdot x\cdot \frac{1}{x}\geq4$$ Hence $x+y\geq 2$. Also for $x=y=1$, you have $x+y=2$, hence $2$ indeed the minimum value.