Is the Geometric mean closer to the Arithmetic or Harmonic mean on the number line?

Question

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?

Answer 1

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

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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

Answer 2

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 $a = b.$