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

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$$a+\frac1a\geq2 \Longleftrightarrow the square of any real number is non-negative so we have (a-1)^2\geq0 (don't understand this part) \Longleftrightarrow a^2-2a+1\geq0 \Longleftrightarrow a^2+1\geq2a \Longleftrightarrow since a>0 then so is a+\frac1a≥2 if a>0 ## 1 Answer Think about it in the other direction: If you square any real number you get a nonnegative result, so$$(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.

• So the second part would be: 0≤(a-1)^2-->-a-(1/a)≤-2 – SpankyS Jan 31 '14 at 1:08
• Has this inequality got the official name? – Andrew Bumetsov Apr 26 at 8:58