Proving the AM-GM Inequality?

Question

Could anyone provide a simple, easily digestible, and well-explained proof of the AM-GM inequality (the multi-value version)?

Much appreciated.

Answer 1

Let $a$ and $b$ be positive numbers. Start with the fact that $(a-b)^2>0$ unless $a=b$. Expand the square to see that $a^2-2ab+b^2>0$. Add $4ab$ to both sides: $a^2+2ab+b^2>4ab$. This is the same as $(a+b)^2>4ab$. Finally, take a square root: $a+b>2\sqrt{ab}$, whence ${a+b\over 2}>\sqrt{ab}$, unless $a=b$.

Answer 2

I have a good answer for this !

First, let me prove the result that if there are two numbers of the same sum, the numbers which are closer together have a greater product.

$$ t + u = x + y $$

Let $t,u$ be further apart from each other than $x,y$

But, $$ t - u \gt x - y $$

Consider the following identity, writing the product of two numbers as a function of it's sum and difference. Since we are subtracting a positive quantity, and the sum is the same, the greater product is of the numbers with the lesser difference.

$$4ab = (a+b)^{2} - (a-b)^{2}$$

From this, we clearly see that $$xy \gt tu$$

Now, let's consider a set of numbers

$$a, b, c {\dots} n$$

Let us choose its arithmetic mean, $M$ given by

$$M = \frac{a+b+c+{\dots}n}{n}$$

The product of this series is

$$P = a.b.c.{\dots}n$$

Now, we make another series of numbers $$a_{1}, b_{1}, c_{1} {\dots} n_{1}$$

We choose this series in such a way that all elements are equal but we make one change. We choose one number less than $M$, (say $a_{1}$) and write $a_{1} = M$.

However, we'd like to preserve the sum and the $AM$ of this series so we'd choose a number greater than $M$, say $b_{1}$ and reduce its value so that $a_{1} + b_{1} = a + b$. Since, $a_{1}$ has been increased, $b_{1}$ must be decreased. All the other values remain as their older counterparts.

$c_{1} = c, d_{1} = d, {\dots} n_{1} = n$

Now we observe,

$P_{1} = a_{1}.b_{1}.c_{1}{\dots}n_{1}$ $=Mb_{1}.c{\dots}n$

Since the product of the terms $c_{1}d_{1}{\dots}n_{1} = cd{\dots}n$, and $Mb_{1} \gt ab$,

$$\implies P_{1} \gt P$$

Now, we construct another series of terms, $a_{2}, b_{2},c_{2}{\dots}n_{2}$ in the same way where all the terms are equal to their counterparts of the previous series. We write, $ b_{2} = M$, and increase the value of a number less than $M$, (say $c_{2}$) so that the sum $b_{2}+c_{2} = b_{1} + c_{1}$ is preserved, but $b_{2}c_{2} \gt b_{1}c_{1}$, since $b_{2},c_{2}$ are closer together.

Now, we observe the product of this series.

$$P_{2}= a_{2}b_{2}c_{2}{\dots}n_{2}$$ $$\implies P_{2} = M^{2}c_{2}{\dots}n_{2}$$

So, by the similar logic, $$P_{2} \gt P_{1}$$

We go on constructing many series making a new element equal to M each time and increasing the product we get,

$$P_{n} \gt {\dots} \gt P_{2} \gt P_{1} \gt P$$ $$\implies P_{n} = M^{n} \gt P$$ $$\implies ( {\frac{a+b+c+\dots+n}{n}})^{n} \gt {a.b.c.{\dots}n}$$

But, both the $L.H.S.$ and the $R.H.S$ are positive quantities.

$$\implies \frac{a+b+c\dots+n}{n} \gt {(a.b.c.{\dots}n)}^{\frac{1}{n}}$$

And, Voila !

$$\implies A.M \gt G.M.$$