How to easily prove $x+\frac{1}{x} \ge 2 \quad ∀x\in ℝ^+$

Question

When I tried to solve some certain math problem (an inequation) for pivate exercise purposes, I had to prove that $x+\frac{1}{x} \ge 2 \quad ∀x\in ℝ^+$, I solved it with tools from differential calculus (prooving that there is a local minimum at $(1,2)$ etc), because this was my only concept. But I guess one can prove this in a much simpler way, but I strangely do not get it — So: How can one prove this the most effective way?

Answer 1

Hint

Deduce the desired inequality from $$(x-1)^2\ge0$$

Answer 2

Hint: since $x$ is positive, multiply both sides by $x$. Then subtract $2x$ from both sides. Now factor to find... and now just reverse your steps :)

Answer 3

$$0 \leq \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2$$

Answer 4

$$x+ \frac{1}{x} \geq 2 \iff x^2+1 \geq 2x \iff x^2-2x+1 \geq \iff (x-1)^2 \geq 0$$

$$\text{The latter is true since $z^2=0 \iff z=0$} $$.