<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] $$(x-1)^2\geq  0\\ \Longrightarrow x^2-2x+1\geq 0\\ \Longrightarrow x^2+1\geq2x$$ 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 $$(x-1)^2\geq  0$$
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