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.


$\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.

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  • $\begingroup$ 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

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