Simplifying x + 1/x < 2

Question

I can't seem to find where my algebra is wrong in my assumption that x + 1/x < 2 can be simplified to (x-1)^2 < 0.

I start with x in the set of real numbers such that x + 1/x < 2 so I then subtract x from both sides leading to

1/x < 2 - x

I then multiply x to both sides to give me

1 < 2x-x^2

and then subtract one

0 < -x^2 +2x -1

I then add one to the other side and subtract 2x and add x^2 to get

x^2 -2x + 1 < 0

and then factor and simplify to

(x-1)^2 < 0

I am simply not seeing where I am breaking rules. I understand how to simplify this inequality I am more wondering why what I did in the above steps is invalid.

It has been a very long time since I have dealt with inequalities so that may be why I am not seeing what is probably an obvious mistake. Any help with this is greatly appreciated.

Edit: x is a real number and I understand that (x-c)^2 >=0 for all x and c in Real Numbers but I am wondering where my algebra went wrong to lead me to this contradiction.

Answer 1

Despite the existing solutions, since OP asks for why the calculations are wrong, I am going to write a few lines.

$$\frac1x < 2 - x \quad(\text{correct step})$$ I then multiply $x$ to both sides to give me $$1 < 2x-x^2.\quad(\text{incorrect deduction})$$

The logic is that the inequality is preserved when both sides are multiplied by a positive number; while it is reversed when they are multiplied by a negative number. $$a>0 \text{ and } b<c \implies ab<ac \\ a<0 \text{ and } b<c \implies ab>ac$$ You have to divide two cases. $$\begin{cases}1 < 2x-x^2 \qquad x>0 \\ 1>2x-x^2 \qquad x<0\end{cases}$$ Rewrite them as squares. $$\begin{cases}-(x-1)^2 < 0 \qquad x>0 \text{ rejected} \\ (x-1)^2 \qquad x<0 \text{ always true} \end{cases}$$ This allows us to conclude that $x<0$ is the necessary and sufficient condition for the inequality $x+1/x<2$.

Answer 2

It's not equivalent to $(x-1)^2<0$ because $x$ can be negative.

We need to make following steps. $$x+\frac{1}{x}<2$$ it's $$x+\frac{1}{x}-2<0$$ or $$\frac{x^2-2x+1}{x}<0$$ or $$\frac{(x-1)^2}{x}<0.$$ If we need to solve this inequality then we obtain $x<0.$

Answer 3

Note that from AM GM, for all positive $x$, $$x+\frac{1}{x} \ge 2$$

Thus the given inequality can only be satisfied when $x < 0$

Answer 4

Note that when multiplying an inequality throughout by something (whether it be a variable or a constant), you need to consider its sign.

Here, $x$ is a real variable, so we have to consider cases. We can rule out $x = 0$ as it leaves one term on the LHS undefined. So just consider $x>0$ and $x<0$.

For the first case, the inequality does not change direction, leaving us with $(x-1)^2<0$. Obviously this has no real solution, so we can conclude that $x>0$ cannot satisfy the original inequality.

For the second case, the inequality changes direction, giving us $(x-1)^2>0$, which holds trivially true for all real $x$. This means that all $x<0$ satisfy the original inequality.