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**A geometric proof for the Arithmetic and Geometric Mean**

There is more than one way to define the mean of a number. The arithmetic mean is the mean we learn at secondary school – for 2 numbers a and b it is:

(a + b) /2.

The geometric mean on the other hand is defined as:

(x_{1}.x_{2}.x_{3}…x_{n})^{1/n}

So for example with the numbers 1,2,3 the geometric mean is (1 x 2 x 3)^{1/3}.

With 2 numbers, a and b, the geometric mean is (ab)^{1/2}.

We can then use the above diagram to prove that (a + b) /2 ≥ (ab)^{1/2} for all a and b. Indeed this inequality holds more generally and it can be proved that the Arithmetic mean ≥ Geometric mean.

Step (1) We draw a triangle as above, with the line MQ a diameter, and therefore angle MNQ a right angle (from the circle theorems). Let MP be the length a, and let PQ be the length b.

Step (2) We can find the length of the green line OR, because this is the radius of the circle. Given that the length a+b was the diameter, then (a+b) /2 is the radius.

Step (3) We then attempt to find an equation for the length of the purple line PN.

We find MN using Pythagoras: (MN)^{2} = a^{2} +x^{2}

We find NQ using Pythagoras: (NQ)^{2} = b^{2} +x^{2}

Therefore the length MQ can also be found by Pythagoras:

(MQ)^{2} = (MN)^{2 } + (NQ)^{2}

(MQ)^{2 } = a^{2} +x^{2} + b^{2} +x^{2}

But MQ = a + b. Therefore:

(a + b)^{2 } = a^{2} +x^{2} + b^{2} +x^{2}

a^{2}+ b^{2} + 2ab = a^{2} +x^{2} + b^{2} +x^{2}

2ab = x^{2} +x^{2}

ab = x^{2}

x = (ab)^{1/2}

Therefore our green line represents the arithmetic mean of 2 numbers (a+b) /2 and our purple line represents the geometric mean of 2 numbers (ab)^{1/2}. The green line will always be greater than the purple line (except when a = b which gives equality) therefore we have a geometrical proof of our inequality.

There is a more rigorous proof of the general case using induction you may wish to explore as well.