How to prove that for all real number $x$, $x+ \frac{1}{x} \geq 2$

Question

Prove that for all real number $x$, $x+ \frac{1}{x} \geq 2$

Although not a formal proof, in terms of approaching the problem could I just manipulate inequalities like this:

$$ x+ \frac{1}{x} \geq 2$$

$$x+ \frac{1}{2} \geq x$$

$$\frac{1}{2} \geq 0$$

Answer 1

I think this is not true, try $x=-1$, but for $x>0$ the inequality is true.

Answer 2

Expand $(\sqrt{x}-\frac{1}{\sqrt{x}})^2 \geq 0$

Answer 3

You can prove it as follows:

$$x + \frac{1}{x} \geq 2$$

$$x^2 -2x +1\geq 0$$

Now, consider $f(x)=x^2-2x+1$. We know it is a parabola. Then, when $x \to \infty$ and $x \to -\infty$, $f(x) \to \infty$. Then, if we check where this function has its minimum, we can check what you want. We know that its minimum is in

$$x=\frac{-b}{2a}$$

$$y=c-\frac{b^2}{4a}$$

That is $(1,0)$. Hence, $f(x)$ has value minimum $0$. Hence, the inequality holds.

Answer 4

Via Fermat's Library twitter status:

A visual proof of $x+ \displaystyle\frac{1}{x} \geq 2$.

Also, the statement is not valid for non-positive real numbers.

Answer 5

The result is not true for all real number.

For positive number, this is true due to $AM-GM$ inequality

$$\frac{x+\frac1x}2\ge \sqrt{x \cdot \frac1x}=1$$