Prove! that $a+(1/a) ≥ 2$ and $a+(1/a) ≤2$

Question

Let $a \in R$

1. If $a>0$, then $a+\frac1a\geq2$
2. 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 $a>0$

Answer 1

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.

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

Answer 2

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 $$a + \frac {1}{a}\le -2$$