<s>[INST] prove：$\left|x+\frac{1}{x}\right|\geq 2$ prove：
$\left|x+\frac{1}{x}\right|\geq 2$
Can I just use 
$\left|\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2\right|\geq 2$
and
$\left|\left(\sqrt{-x}\right)^2+\left(\frac{1}{\sqrt{-x}}\right)^2\right|\geq 2$?
 [/INST] $$
a^2 - 2ab + b^2 = (a-b)^2
$$
$$
x + \frac1x \text{ is the same as }\sqrt{x}^2 + {\sqrt{\frac1x}}^2
$$
but there's no "$-2ab$" term in the middle.  Since $a=\sqrt{x}$ and $b=\sqrt{\frac1x}$, we conclude that $-2ab=-2\sqrt{x}\cdot\sqrt{\frac1x}$.  This simplifies to just $-2$.
So
$$
\sqrt{x}^2 -2 + {\sqrt{\frac1x}}^2
$$
is a perfect square: $a^2-2ab+b^2=(a-b)^2=\left(\sqrt{x}-\sqrt{\frac1x}\right)^2=x-2+\frac1x$.
So
$$
x+\frac1x = \left(x-2+\frac1x\right) + 2 = \left(\sqrt{x}-\sqrt{\frac1x}\right)^2 + 2 = \underbrace{\text{a square} + 2}.
$$
Since it's a square plus $2$, it is $\ge 2$.
(All this assumes $x$ is positive; it $x$ is negative, then it's not true that $x+\frac1x\ge 2$.)
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