IIT JEE , AIEEE Forum - Mathematics , Algebra

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Proof by Cauchy :

Case : All terms are equal

If all the terms are equal:

$x_1 = x_2 = \cdots = x_n$

then their sum is nx₁, so their arithmetic mean is x₁; and their product is x₁ⁿ, so their geometric mean is x₁; therefore, the arithmetic mean and geometric mean are equal, as desired.

Case : All terms are not equal

It remains to show that if not all the terms are equal, then the arithmetic mean is greater than the geometric mean. Clearly, this is only possible when n > 1.

Subcase : When n = 2

If n = 2, then we have two terms, x₁ and x₂, and since (by our assumption) not all terms are equal, we have:

$x_1\,\;$	$\ne x_2\,\;$
$x_1 - x_2\,\;$	$\ne 0\,\;$
$\left( x_1 - x_2 \right) ^2\,\;$	$> 0\,\;$
$x_1^2 - 2 x_1 x_2 + x_2^2\,\;$	$> 0\,\;$
$x_1^2 + 2 x_1 x_2 + x_2^2\,\;$	$> 4 x_1 x_2\,\;$
$\left( x_1 + x_2 \right) ^2\,\;$	$> 4 x_1 x_2\,\;$
$\left( \frac{x_1 + x_2}{2} \right)^2\,\;$	$> x_1 x_2\,\;$
$\frac{x_1 + x_2}{2}\,\;$	$> \sqrt{x_1 x_2}\,\;$

Subcase : When n = 2^k

We proceed by mathematical induction.

In the base case, k = 1, so n = 2. We have already shown that the inequality holds where n = 2, so we are done.

Now, suppose that for a given k > 1, we have already shown that the inequality holds for n = 2^k?1, and we wish to show that it holds for n = 2^k. To do so, we proceed as follows:

$\frac{x_1 + x_2 + \cdots + x_{2^k}}{2^k}$	$=\frac{\frac{x_1 + x_2 + \cdots + x_{2^{k-1}}}{2^{k-1}} + \frac{x_{2^{k-1} + 1} + x_{2^{k-1} + 2} + \cdots + x_{2^k}}{2^{k-1}}}{2}$

	$\ge \frac{\sqrt[2^{k-1}]{x_1 \cdot x_2 \cdots x_{2^{k-1}}} + \sqrt[2^{k-1}]{x_{2^{k-1} + 1} \cdot x_{2^{k-1} + 2} \cdots x_{2^k}}}{2}$

	$\ge \sqrt{\sqrt[2^{k-1}]{x_1 \cdot x_2 \cdots x_{2^{k-1}}} \cdot \sqrt[2^{k-1}]{x_{2^{k-1} + 1} \cdot x_{2^{k-1} + 2} \cdots x_{2^k}}}$

	$= \sqrt[2^k]{x_1 \cdot x_2 \cdots x_{2^k}}$

where in the first inequality, the two sides are only equal if both of the following are true:

$x_1 = x_2 = \cdots = x_{2^{k-1}}$

$x_{2^{k-1}+1} = x_{2^{k-1}+2} = \cdots = x_{2^k}$

(in which case the first arithmetic mean and first geometric mean are both equal to x₁, and similarly with the second arithmetic mean and second geometric mean); and in the second inequality, the two sides are only equal if the two geometric means are equal. Since not all 2^k numbers are equal, it's not possible for both inequalities to be equalities, so we know that:

$\frac{x_1 + x_2 + \cdots + x_{2^k}}{2^k} > \sqrt[2^k]{x_1 \cdot x_2 \cdots x_{2^k}}$

Subcase : When n < 2^k

If n is not a natural power of 2, then it is certainly less than some natural power of 2, since the sequence 2, 4, 8, . . ., 2^k, . . . is unbounded above. Therefore, without loss of generality, let m be some natural power of 2 that is greater than n.

So, if we have n terms, then let us denote their arithmetic mean by ?, and expand our list of terms thus:

$x_{n+1} = x_{n+2} = \cdots = x_m = \alpha.$

We then have:

$\alpha\,\;$	$= \frac{x_1 + x_2 + \cdots + x_n}{n}$

	$= \frac{\frac{m}{n} \left( x_1 + x_2 + \cdots + x_n \right)}{m}$

	$= \frac{x_1 + x_2 + \cdots + x_n + \frac{m-n}{n} \left( x_1 + x_2 + \cdots + x_n \right)}{m}$

	$= \frac{x_1 + x_2 + \cdots + x_n + \left( m-n \right) \alpha}{m}$

	$= \frac{x_1 + x_2 + \cdots + x_n + x_{n+1} + \cdots + x_m}{m}$

	$> \sqrt[m]{x_1 \cdot x_2 \cdots x_n \cdot x_{n+1} \cdots x_m}$

	$= \sqrt[m]{x_1 \cdot x_2 \cdots x_n \cdot \alpha^{m-n}},$
so
$\alpha^m\,\;$	$> x_1 \cdot x_2 \cdots x_n \cdot \alpha^{m-n}$
$\alpha^n\,\;$	$> x_1 \cdot x_2 \cdots x_n$
$\alpha\,\;$	$> \sqrt[n]{x_1 \cdot x_2 \cdots x_n}.$

as desired.

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