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For any 2 distinct positive real numbers a and b

Arithmetic mean:A=$\frac{a+b} {2}$

Geometric mean:G=$\sqrt{ab} $

Harmonic mean:H=$\frac{2ab}{a+b}$

Its well known that A>G>H here.

On the number line they look like such:

a----H-G--A--------b

I tried playing around a bit on Desmos and found that the geometric mean was always closer to the harmonic mean than the arithmetic mean.

What are the ways I could prove this in, either intuitively or otherwise?

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A rigorous way is as such

Let y=2$\sqrt{ab}-\frac{2ab}{a+b}-\frac{a+b} {2}$

This can be simplified as $\frac{4(a+b)\sqrt{ab}-4ab-2(a+b)^2}{2(a+b)}$

This further simplifies into $-\frac{(\sqrt{a} +\sqrt{b})^4}{a+b}$

This is always negative as we are given a$\neq$b and a, b>0

So we can say that

$G<\frac{A+H}{2} $

So G lies to the left of the midpoint of A and H meaning G is closer to H than A


Another method which is not fully mine but the concept has been required to solved a question from IITJEE 2007 Mathematics.

Let A$_{1}$,G$_{1}$,H$_{1}$ be the arithmetic, geometric and Harmonic means of a and b.

For n$\geq$ 2 let A$_{n-1}$, H$_{n-1}$ have Arithmetic, geometric and Harmonic means as A$_{n}$, G$_{n}$, H$_{n}$

We see that $$A_{n}=\frac{A_{n-1}+H_{n-1}} {2}$$, $$H_{n}=\frac{2A_{n-1}H_{n-1}}{A_{n-1}+H_{n-1}}$$ and $$G_ {n} =\sqrt{A_{n-1}H_{n-1}}$$

We observe that G$_{n+1}$=G$_{n}$

So the position of the geometric mean does not change.

However applying AM, GM, HM inequality the geometric mean must be less than the average (AM) of Arithmetic and Harmonic means which yet again proves the statement

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By the A.M.-G.M. inequality applied to $A$ and $H,$ $$ G = \sqrt{ab} = \sqrt{\frac{a + b}2\cdot\frac{2ab}{a + b}} = \sqrt{AH} \leqslant \frac{A + H}2, $$ therefore $$ G - H \leqslant A - G, $$ with equality if and only if $A = H$, i.e., if and only if $a = b.$

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  • $\begingroup$ I think its better to justify why geometric mean of a and b is same as the geomtric mean of their arithmetic and harmonic means? $\endgroup$ – Vamsi Krishna Jun 11 at 2:47
  • $\begingroup$ OK, have done so. On another point: $a$ and $b$ were stipulated in the question to be unequal, but I allowed the equality case. I've left it as it is, not wanting to fuss too much, but it might have been better to change the non-strict inequality signs to strict ones and leave out the last line. $\endgroup$ – Calum Gilhooley Jun 11 at 9:30

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