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Randomly thought about this in bio class today, did some quick work. Is this identity known and does it have any applications?

  • Let $\text{am}(a,b)$ denote the arithmetic mean of two numbers.

  • Let $\text{gm}(a,b)$ denote the geometric mean of two numbers.

Consider two arbitrary numbers $x$ and $y$.

Lemma: $\text{am}(\text{am}(x,y),\text{gm}(x,y))=[\text{am}(\sqrt x,\sqrt y)]^2$

Proof:

We have the following two identities:

  • $\text{am}(x,y)=\frac{x+y}{2}$

  • $\text{gm}(x,y)=\sqrt{xy}$

$$\implies\text{am}\left(\frac{x+y}{2},\sqrt{xy}\right)=[\text{am}(\sqrt x,\sqrt y)]^2$$

$$\implies\frac{\frac{x+y}{2}+\sqrt{xy}}{2}=\left[\frac{\sqrt x+\sqrt y}{2}\right]^2$$

$$\implies\frac{x+y+2\sqrt{xy}}{4}=\frac{x+y+2\sqrt{xy}}{4}$$

This completes the proof.

Leonidas Lanier
  • 4,149
  • AM GM inequality is known by almost everyone – Saketh Malyala Jun 07 '17 at 20:56
  • it is applicable in many problems where the minimum of a multivariable expression is needed under a certain domain – Saketh Malyala Jun 07 '17 at 20:56
  • The notion of combining arithmetic and geometric means is covered here. $\text{am}(\text{am}(x,y),\text{gm}(x,y))$ is called $a_1$ in the article. $g_1$ would be $\text{gm}(\text{am}(x,y),\text{gm}(x,y))$ – Χpẘ Jun 07 '17 at 23:57

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