Prove the inequality $x+\frac{1}{x}\geq 2$? I'm reading Courant's Calculus. There is an exercise: Prove the inequality $x+\frac{1}{x}\geq 2$. 
I did the following: For the sake of curiosity, I did:
$$x+\frac{1}{x}\geq 2$$
$$x^2+1\geq 2x$$
$$x^2-2x +1\geq 0$$
And this is just $(x-1)^2$. Well, as the square of a real number is $\geq0$, then:
$$(x-1)^2 \geq 0$$
$$x^2-2x+1 \geq 0$$
$$x^2+1 \geq 2x$$
$$x+\frac{1}{x} \geq 2$$
Which is what was to be proved. Is it correct?

By AM-GM, for all $x>0$:
$$\frac{x+\frac{1}{x}}{2} \geq\sqrt{x\cdot\frac{1}{x}} $$
$$x+\frac{1}{x} \geq2$$
