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Consider two positive real numbers $\frac{1}{a}$ and $\frac{1}{b}$. Show that the 'mean' $$\dfrac{1}{\frac{a+b}{2}}=\dfrac{2}{a+b}$$ will always be less than the arithmetic mean $$\dfrac{a+b}{2ab}.$$

Proving this is not difficult, but I was wondering if there was a name for this inequality, or if it demonstrates another known inequality in action, much like how many known inequalities are simple consequences of say the AM/GM Inequality.

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    $\begingroup$ It is probably more insightful if you replace $\frac 1a$ and $\frac 1b$ by $a$ and $b$. $\endgroup$ – Arnaud Mortier Aug 24 '18 at 6:55
  • $\begingroup$ @ArnaudMortier If we looked at $a$ instead of $1/a$, we would explicitly have to include a non-$0$ criterion. There’s a tradeoff. $\endgroup$ – gen-z ready to perish Aug 24 '18 at 7:18
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    $\begingroup$ @ChaseRyanTaylor since the numbers have to be positive either way, that's not an issue, at least in English. $\endgroup$ – Especially Lime Aug 24 '18 at 7:29
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Yes this is the HM-AM inequality and it is a consequence of AM-GM.

Take a look here

RMS-AM-GM-HM inequalities

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    $\begingroup$ HM is for Harmonic Mean. $\endgroup$ – mathreadler Aug 24 '18 at 7:07
  • $\begingroup$ @mathreadler Yes exactly! $\endgroup$ – user Aug 24 '18 at 7:20
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This is just saying that the inverse of the mean is at most the mean of the inverses:

$$\frac{1}{\frac{a+b}2} \leq \frac{\frac{1}{a} + \frac{1}{b}} 2$$

This follows from the convexity of the function $\frac 1 x$ (on the positive real numbers). Of course, convexity is closely related to the AM-GM inequality, so maybe it's not surprising that this can also been seen as related to AM-GM.

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