Is there a name for this inequality?

Question

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.

Answer 1

Yes this is the HM-AM inequality and it is a consequence of AM-GM.

Take a look here

RMS-AM-GM-HM inequalities

Answer 2

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.