How to solve inequality $1-\frac{1}{x}\geq 0$ correctly? When I try to solve this inequality I get the wrong results:
$$1-\frac{1}{x}\geq 0$$
Assuming $x\neq 0$
It's really embarrassing that I can't see what I'm doing wrong.
So please help me, telling me where the maths goes wrong :D

$$1-\frac{1}{x}\geq 0 \Leftrightarrow 1\geq\frac{1}{x} \Leftrightarrow x\geq1$$
But this is wrong, since I can easily see $1-\frac{1}{x}\geq 0$ when x<0. So what am I doing wong?
 A: As an alternative, we have that
$$1-\frac{1}{x}\geq 0 \iff \frac1x\le 1$$
and

*

*for $x>0\quad x\cdot \frac1x\le x\cdot1 \implies x\ge 1$

*for $x<0\quad x\cdot \frac1x\ge x\cdot1  \implies x\le 1$ that is $x<0$
Also a graphical check for this simple inequality may help

A: It's $$\frac{x-1}{x}\geq0,$$ which gives $$[1,+\infty)\cup(-\infty,0).$$
We used the intervals method.
In the expression $(x-a)^k$ the natural number $k$ is named a degree of the point $a$.
We need to solve
$$\frac{(x-1)^1}{(x-0)^1}\geq0.$$
The degree of $1$ is $1$ and the degree of $0$ is $1$.
If we go through a point $1$ so the sign of $x-1$ is changed because the degree of this point is odd.
If we go through a point $0$ so the sign of $x-0$ is changed because the degree of this point is odd.
For $x>1$ the expression $\frac{x-1}{x}$ has a sign $+$.
Thus, we obtain the following signs: $++++0----1++++$.
It's better to draw an $x$-axis and to put there points $0$ and $1$, which gives the following sequence of sings: $+$$-$$+$ and we obtain the answer: $[1,+\infty)\cup(-\infty,0).$
Another example.
Let we need to solve $\frac{(x-2)^2}{(x-1)(x-3)}\geq0.$
The degree of $2$ is even (the sign is not changed) and degrees of $1$ and $3$ are odds (the sign is changed).
Also, the sign of the expression  $\frac{(x-2)^2}{(x-1)(x-3)}$ for $x>3$ is $+$.
Thus, we have the following sequence of sings: $$+--+,$$ which gives the answer:
$$(-\infty,1)\cup(3,+\infty)\cup\{2\}$$
A: Set apart forbidden value $x=0$, you are with a quotient:
$$\dfrac{x-1}{x}>0$$
Here is a very convenient "recipe": the sign of the quotient $N/D$ is the same as the sign of the product $N*D=x(x-1)$.
As this product is a quadratic beginning with a positive coefficient for $x^2$, with real roots $0,1$, we can conclude that this product is $>0$ outside the interval $[0,1]$ delimited by these roots.
A: What you did wrong was the last double-implication.
In general, you can turn
$$
A \le C
$$
into
$$
AB \le CB
$$
only if you know that $B \ge 0$. When $B < 0$,
$$
A \le C
$$
is equivalent to
$$
AB \ge CB
$$
instead!
For example, we all agree that
$$
1 \le 4,
$$
right? Now multiply both sides by $-1$, and you get
$$
-1 ??? -4
$$
What should replace the "???" Answer: a $\ge$ sign, not a $\le$ sign!
Back to your equation. You had
$$
1 \ge \frac{1}{x} 
$$
and you wanted to multiply by $x$ ... but depending on whether $x$ is positive or negative, the result will be different.
So split into two cases: You say
"In the case where $x > 0$, this is equivalent to
$$
x \ge 1
$$
so we get, as solutions, all positive numbers greater than or equal to $1$.
In the case where we look for solutions $x$ with $x < 0$, our equation is equivalent (after multiplying both sides by $x$, a negative number!) to
$$
x \le 1.
$$
Of course, not all solutions of this equation satisfy our assumptions for this part of the work: numbers like $\frac{1}{2}$ are less than $1$, so they satisfy the equation, but they're not negative, so they don't satisfy the assumptions. Which negative numbers are less than or equal to $1$? All of them.
So the solutions overall are

*

*All negative numbers, and

*All positive numbers greater than or equal to $1$.

