4
$\begingroup$

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 $a>0$

$\endgroup$
9
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.