Prove $x + \frac{1}{x} \geq 2$ for $x>0$. Proof that $x+\frac{1}{x}\geq2$ for $x>0$
Would this be correct?
$x*(x+\frac{1}{x}\geq2)$
$x^2+1\geq2x$
$x^2-2x+1\geq2x-2x$
$x^2-2x+1\geq0$
Plug in 1 for x:
$(1)^2-2(1)+1\geq0$
$1-2+1\geq0$
$0\geq0$ Therefore, $x+\frac{1}{2}\geq2$ is true for $x>0$

We know for positive $a,b$, $\frac{ a + b}{2} \geq \sqrt{ab} $. Put $ a = x^2 $ and $b = 1$ and we obtain 
$$ x + \frac{1}{x} \geq 2 $$
Added: You can also use calculus: Put $f(x) = x + \frac{1}{x} $. we have $f'(x) = 1 - \frac{1}{x^2} $ with critical values (points where $f'$ vanishes ) : $x=\pm1$. It is easy to see that $x = 1 $ will furnish a global minimum. Hence $f(x) \geq f(1) $ for all $x > 0 $. It follows that 
$$ x + \frac{1}{x} \geq 2 $$
