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$$a+\frac1a\geq2$ 
$\Longleftrightarrow$ the square of any real number is non-negative
  so we have $(a-1)^2\geq0$ (don't understand this part)
$\Longleftrightarrow$ $a^2-2a+1\geq0$
$\Longleftrightarrow$ $a^2+1\geq2a$
$\Longleftrightarrow$ since $a>0$ then so is $
  a+\frac1a≥2$ if $a>0$
 A: Think about it in the other direction: If you square any real number you get a nonnegative result, so
$$(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.
A: Since the above answer already gave you the proof for first one, I'll discuss the second inequality
$$a+\frac1a\leq2$$
This is a wrong inequality, you'll understand that in a moment.
Let's start with the first inequality user61527 used: $$(a - 1)^2 \ge 0$$ This false for $a \lt 0$ because $(a - 1)^2$ can never be $0$ for any negative value. It will be zero only at $a=1$ which is out of domain for $a$ as $a\lt 0$.
So we'll use the following inequality $$(a + 1)^2\ge 0$$ The equality will hold at $a=-1$, now this fits in fomain of $a$
$$a^2 + 2a + 1 \ge 0$$
Dividing by $a$ will alter the inequality since it is a negative number
$$a + 2 + \frac {1}{a}\le 0$$
So the actual inequation for $a\lt 0$ should be $$a + \frac {1}{a}\le -2$$
