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
$\left|\left(\sqrt{-x}\right)^2+\left(\frac{1}{\sqrt{-x}}\right)^2\right|\geq 2$?
 A: Hint: Prove that $$\left(x+\frac{1}{x}\right)^2 \geq 4$$
A: Without loss of genarality we can consider $x\ge 0$(Reason: $x$ and $\frac{1}{x}$ has same sign so if $x<0$ the in place of $x$ we will take $-x(>0)$ and it will not make any difference since $\left| x+\frac{1}{x}\right |=\left| -x+\frac{1}{-x}\right |$)
Then we have,
$\left| x+\frac{1}{x}\right |=x+\frac{1}{x}=(\frac{1}{\sqrt{x}}-\sqrt{x})^2+2\ge 2$
You can use(with its proof which is done by my above arguement) either one(not both ) of the following(if you want to handle only real stuff) according to $x$ is positive or negative,
$\left|\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2\right|\geq 2$ or $\left|\left(\sqrt{-x}\right)^2+\left(\frac{1}{\sqrt{-x}}\right)^2\right|\geq 2$
You cant use both since ,if $x$ is positive then $\sqrt{-x}$ is complex.And if $x<0$ then $\sqrt{x}$ is complex.
A: Find an absolute minimum:
$$\frac{d}{dx}\left|x+\frac{1}{x}\right|=\left(1-\frac{1}{x^2}\right)\frac{|x+\frac{1}{x}|}{x+\frac{1}{x}}$$
$$0=1-\frac{1}{x^2}$$
$$x=\pm{1}$$
$$|x+\frac{1}{x}|_{x=\pm{1}}=2$$
Therefore $2$ is a local minimum.
It can be demonstrated that $(\pm{1},2)$ is the absolute minimum by observing the boundaries of the intervals $(0,\infty)$ and $(-\infty,0)$.
$$\lim_{x\to{-\infty}}|x+\frac{1}{x}|=\infty$$
$$\lim_{x\to{0^-}}|x+\frac{1}{x}|=\infty$$
$$\lim_{x\to{0^+}}|x+\frac{1}{x}|=\infty$$
$$\lim_{x\to{\infty}}|x+\frac{1}{x}|=\infty$$
Therefore $(\pm{1},2)$ is the absolute minimum.
A: For $x\neq 0$, we have $(x-1)^{2}\geq 0$. Therefore, $x^{2}-2x+1\geq 0$ and hence $x^{2}+1\geq 2x$. If $x>0$, then we are done. If $x< 0$, then $-x> 0$. Hence, 
$\displaystyle x^{2}+1\geq 2(-x)\implies -x-\frac{1}{x}\geq 2\implies x+\frac{1}{x}\leq -2$.
A: $$
a^2 - 2ab + b^2 = (a-b)^2
$$
$$
x + \frac1x \text{ is the same as }\sqrt{x}^2 + {\sqrt{\frac1x}}^2
$$
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 $x+\frac1x\ge 2$.)
A: Denote $x+\frac{1}{x}=v$. 
Obviously $|v|>2$ if either $v>2$ or $-v<-2$. The first case is therefore $x +\frac{1}{x}>2$ and it becomes $(x-1)^2>0$, which is always true. The second case is $-x-\frac{1}{x}<-2$ or $\frac{x^2-2x+1}{x}>0$ and also breaks down into 2 cases: either $(x-1)^2>0 \cap x>0$ or $(x-1)^2<0 \cap x<0$. Obviously the second case is never true and the first case is always true. 
Hence $|x+\frac{1}{x}|$ is always larger than $2$.
A: If $x>0$, then by AM-GM inequality, $x+1/x \geq 2\cdot \sqrt{x\cdot(1/x)}=2$, hence $|x+1/x| \geq 2$. If $x<0$, then $-x>0$ and therefore $|x+1/x|=|-x-1/x|\geq 2$ by the former result.
