How to prove this inequality $ x + \frac{1}{x} \geq 2 $

I was asked to prove that:

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

for all values of $ x > 0 $

I tried substituting random numbers into $x$ and I did get the answer greater than $2$. But I have a feeling that this is an unprofessional way of proving this. So how do I prove this inequality?

For $x\gt 0$ you have $$x+\frac{1}{x}-2 = \frac{x^2}x+\frac1x-\frac{2x}x= \frac{x^2-2x+1}{x} = \frac{\left(x-1\right)^2}{x} \geq 0 $$

This post is due to the reason that no-one elaborated the AM-GM technique, and any beginner not knowing this method might get help to learn this method of proof.

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

Please note that this is a direct consequence of- a perfect square is always postive.

Since $x\gt 0$ you can multiply through by $x$ to clear fractions without changing the sense of the inequality. This gives $$x^2+1\ge2x$$Subtract $2x$ from each side:$$x^2-2x+1\ge0$$ or $$(x-1)^2\ge0$$ Which is true, with equality only if $x=1$ since squares are non-negative.

Now note that each of these steps can be reversed to take us from the last statement, which we know to be true, to the statement in your question, which you want to prove. Since $x\gt0$ you can divide by $x$. Write it out carefully and you will have your proof.