Arithmetic mean greater than equal to Geometric mean condition Let's say there are two numbers a and b: 
In all the books it is said that the arithmetic mean of a and b is greater than or equal to the geometric mean of $a$ and $b$ if $a>0, b>0$. 
But why can't we say the arithmetic mean of a and b is greater than or equal to the geometric mean of $a$ and $b$ if $a>=0, b>=0$?   
 A: 
But why can't we say the arithmetic mean of $a$ and $b$ is greater than or equal to the geometric mean of $a$ and $b$ if $a\ge 0,\,b\ge 0$?

Well we can say that too, since it's true. For if $a\ge 0,\,b\ge 0,$ then we may write $a=x^2,\,b=y^2.$ Now since $(x-y)^2=x^2+y^2-2xy\ge 0,$ or that $x^2+y^2\ge 2xy,$ it follows that $$\frac{x^2+y^2}{2}\ge xy,$$ or in other terms, that $$\frac{a+b}{2}\ge \sqrt a\sqrt b=\sqrt{ab},$$ provided we choose $x,y\ge 0,$ which we do.
A: $a=0$ or $b=0$ is a trivial condition for the AM-GM inequality as if any one of them is 0, GM becomes 0 and AM will obviously be greater than 0. $a=b=0$ is the equality condition for the equality. Hence the books display it as $a{\gt}0$ and $b{\gt}0$ as the non trivial condition. 
A: Let $a\ge 0$ and $b\ge 0$ be two reals.
we can write
$$a=(\sqrt{a})^2 \text{ and } \; b=(\sqrt{b})^2$$
thus
$$a+b-2\sqrt{ab}=(\sqrt{a}-\sqrt{b})^2\ge 0$$
and
$$\frac{a+b}{2} \ge \sqrt{ab}$$
In the case $a>b\ge 0$, we have
$$\frac{a+b}{2} > \sqrt{ab}$$
A: Why do you say that we can't say it?     
We can say it. The inequality 
$$\frac{a+b}{2} \ge \sqrt{ab}$$ 
is true for all $a \geq 0, b \geq 0$ 
There we go. We said it and it's a valid statement.   
Side note: maybe it was enough for the purpose of the book/chapter/whatever to claim the statement is true just for positive $a$ and $b$. In general... if you meet some statement A in a book, it does not mean there isn't a more general version B of this statement that is also true. 
