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 $x>0$
 A: 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 
$$ x + \frac{1}{x} \geq 2 $$
A: $$(x-1)^2\geq  0\\ \Longrightarrow x^2-2x+1\geq 0\\ \Longrightarrow x^2+1\geq2x$$ Then we assume $x$ is positive and divide by $x$. $$\Longrightarrow x+\frac1{x}\geq 2$$ See if you can prove it for negative $x$ and the case where $x=0$ then see if you can prove $$(x-1)^2\geq  0$$
A: Hint: You have to prove that 1 is the minimum of the function.
A: Proof by contradiction:
Assume that the hypothesis is wrong. Therefore, $ \exists\ x \gt 0 $ such that; $$ x + \dfrac{1}{x} \lt 2 $$
Now, follow along:
$$\begin{align}
x + \dfrac{1}{x} &\lt 2 \\
x^2 + 1 &\lt 2x \tag{$ \because x \gt 0 $} \\
x^2 - 2x + 1 &\lt 0 \\
(x - 1)^2 &\lt 0
\end{align}$$
which is a contradiction as a perfect square can never be negative for real numbers.
