<s>[INST] 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 $x>0$
 [/INST] 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.
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