Arithmetic Mean Never Less than Harmonic Mean

Theorem

Let $x_1, x_2, \ldots, x_n \in \R$ be real numbers which are all positive.

Let $A_n$ be the arithmetic mean of $x_1, x_2, \ldots, x_n$.

Let $H_n$ be the harmonic mean of $x_1, x_2, \ldots, x_n$.

Then $A_n \ge H_n$.

Proof

The arithmetic mean of $x_1, x_2, \ldots, x_n$ is defined as:

$\displaystyle A_n = \frac 1 n \left({\sum_{k=1}^n x_k}\right)$

The harmonic mean of $x_1, x_2, \ldots, x_n$ is defined as:

$\displaystyle \frac 1 H_n = \frac 1 n \left({\sum_{k=1}^n \frac 1 {x_k}}\right)$

As $\forall k \in \left[{1 .. n}\right]: x_k > 0$, we can express all the $x_k$'s as squares:

$\forall k \in \left[{1 .. n}\right]: x_k = y_k^2$

without affecting the result.

Thus we have:

• $\displaystyle A_n = \frac 1 n\left({\sum_{k=1}^n y_k^2}\right)$

• $\displaystyle \frac 1 H_n = \frac 1 n \left({\sum_{k=1}^n \frac 1 {y_k^2}}\right)$

Now let's see what happens when we multiply them together:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {A_n} {H_n}\)	\(=\)	\(\displaystyle \frac 1 n \left({\sum_{k=1}^n y_k^2}\right) \frac 1 n \left({\sum_{k=1}^n \frac 1 {y_k^2} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\ge\)	\(\displaystyle \frac 1 {n^2} \left({\sum_{k=1}^n \frac {y_k} {y_k} }\right)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Cauchy's Inequality
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {n^2} \left({\sum_{k=1}^n 1}\right)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {n^2} {n^2} = 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So $\dfrac {A_n} {H_n} \ge 1$ and the result follows.

$\blacksquare$

See Also

• Cauchy's Mean Theorem

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 1.12 \ (5)$