Prove! that $a+(1/a) ≥ 2$ and $a+(1/a) ≤2$ Let $a \in R$ 


*

*If $a>0$, then $a+\frac1a\geq2$

*If $a<0$, then $a+\frac1a\leq2$


This is how someone explained the first one to me but still not really sure about it.
Proof:
$\Longleftrightarrow$$a+\frac1a\geq2$ 
$\Longleftrightarrow$ the square of any real number is non-negative
  so we have $(a-1)^2\geq0$ (don't understand this part)
$\Longleftrightarrow$ $a^2-2a+1\geq0$
$\Longleftrightarrow$ $a^2+1\geq2a$
$\Longleftrightarrow$ since $a>0$ then so is $
  a+\frac1a≥2$ if $a>0$

Think about it in the other direction: If you square any real number you get a nonnegative result, so
$$(a - 1)^2 \ge 0$$
Expand the left side:
$$a^2 - 2a + 1 \ge 0$$
If $a > 0$, we divide by $a$ to find
$$a - 2 + \frac 1 a \ge 0$$
or upon rearrangement, the desired inequality.

If $a < 0$, division by $a$ reverses the inequality.
