Is it true that $xy\geq x+y$ for all $x,y\in \mathbb R$? I just see the GM-AM inequality. But I would like to compare $xy$ with $x+y$ for any $(x, y)\in\mathbb{R}^2$.
It looks like $xy>x+y$ since the first one is multiplication and the second one is addition.
Can we prove this? For $(x, y)\in\mathbb{N}^2$, can we prove it now?
 A: We have $xy \geq x+y \iff (x-1)(y-1) \geq 1$.
A: This only holds for a certain subset of $\mathbb R^2$. To find it, solve the equation $xy=x+y$, since this will be the border between $S_1:=\{xy < x+y\}$ and $S_2:=\{xy > x+y\}$.
Your equation is valid only on $\overline{S_2}$
Let us now find $S_2$:
$$xy = x + y \Rightarrow_{x\ne 0} y = 1 + \frac yx \Rightarrow y(1-\frac1x) = 1 \Rightarrow y = \frac1{1-\frac1x}$$
(Fun fact: $y$ is the Hölder conjugate of $x$)
Now for $x=0$ we have $y=0$ implied. $S_2$ is now given by $y > \frac1{1-\frac1x}$ if $x$ is positive and $y < \frac1{1-\frac1x}$ if $x$ is negative. See this for an image of $S_2$

A: No, the first example gives the answer: with $x=y=1$ you have $xy=1$ and $x+y=2$. Same if $x=1$ or $y=1$, or if $(x,y)=(2,2)$. But in the other cases where $(x,y)\in \mathbb{N}^2$ it works, since $xy-(x+y)=(x-1)(y-1)-1> 0$.
A: $$xy\geq x+y
\Rightarrow xy-x-y\geq 0
\Rightarrow xy-x-y+1\geq 1~~ \text{(adding 1 to both sides)}
\Rightarrow x (y-1)-(y-1)\geq 1
\Rightarrow (y-1)(x-1)\geq 1$$
A: For natural numbers it is true if both $x$ and $y$ are at least $2$ or greater and one of them is $3$ or greater. In general it's false.
