Prove the Inequality of Arithmetic and Geometric Means (AM-GM inequality)

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In this video I will show you how to prove the AM-GM inequality. There are other ways of proving this, this is just one way. I hope this helps.

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You can construct the inequality starting from the fact that for two numbers a, b greater than or equal to zero, that (a-b)(a-b) is greater than or equal to zero.

jimpim

Cant a and b be negative? I know you posed the problem as a and b are positive at the beginning but cant a square root of a number be positive and negative therefore a and b can be both negative?

thehardlife

The way to prove such a statement is by starting with an equation that is true for all positive values of a and b.

√a - √b ≥ 0, for a ≥ b [it will always be the case depending on which positive numbers you are referring to]
[Example: Say the numbers are 3 and 5, so a = 5, b = 3]

Then you square both sides:

a + b - 2√ab ≥ 0
a + b ≥ 2√ab

And finally:

(a + b) / 2 ≥ √ab [Proven]

Note that this is just the proof for 2 positive real numbers.

DoubleShaurya

why do you have to suppose that the result is false when a and b are different and different from zero?

veronicaisoo

In general a + b + c + d.... N times all divided by n. You know

nutriascoc

I find the visualization of this to be an easier proof to grasp the meaning.

even bigger picture when you think that RMS >= AM >= GM >= HM.

Cool contradiction LOL! thats hilarious!

thomasjefferson

I have a similar problem to this

Why a+b+c > 3 × ³√abc

AliiSattari

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