# Proof of the Arithmetic Mean - Geometric Mean Inequality

We will prove the Arithmetic Mean-Geometric Mean Inequality using a proof method called Forward-Backward Induction.

To outline the proof, in the forward argument, we will show that the statement is true for larger and larger values of $n$ (specifically for all $n$ powers of $2$). In the backward argument, we will show that if the statement is true for $n$ variables, then it is also true for $n-1$ variables. By combining these arguments, we show that the statement is true for any value of $n$, by first applying the forward argument to show that it is true for a power of 2 that is larger than $n$, and then using the backward argument to show that it is true for $n$.

### AM-GM inequality: For $n$ non-negative real values $x_1, x_2, \ldots x_n$,

$\displaystyle \frac { \displaystyle \sum_{i=1}^n x_i} {n} \geq \sqrt[n] { \prod_{i=1}^n x_i } .$

Proof: (Forward) We will show this by induction. The base case $2^1=2$ is proved in Complete the Square. For the induction step, suppose the statement is true for some $2^k$; we would like to show that the statement is true for $2^{k+1}$. Given $\{a_i\}_{i=1} ^{2^{k+1}}$ positive real values, we divide the set in half (obtaining $\{a_i\}_{i=1} ^{2^{k}}$ and $\{a_i\}_{i=2^k+1} ^{2^{k+1}}$), and then apply the induction hypothesis to each set.

\begin{aligned}{} \frac{a_1+a_2+\ldots+a_{2^{k+1}}}{2^{k+1}} = & \frac {1}{2}\left( \frac{a_1+a_2+\ldots a_{2^k}}{2^k} + \frac {a_{2^k +1} + a_{2^k+2} + \ldots a_{2^{k+1}} } {2^k} \right)\\ \geq & \frac {1}{2} \left( \sqrt[2^k] {a_1 \cdot a_2 \cdots a_{2^k} } + \sqrt[2^k] {a_{2^k+1} \cdot a_{2^k+2} \cdots a_{2^{k+1} } } \right) \\ \geq & \sqrt[2^{k+1}] {a_1 \cdot a_2 \cdots a_{2^k} \cdot a_{2^k+1} \cdot a_{2^k+2} \cdots a_{2^{k+1}} }\\ \end{aligned}

The first inequality follows from using the Induction Hypothesis twice, while the second inequality follows from the 2-variable case, by setting $x_1 = a_1 \cdot a_2 \cdots a_{2^k}$ and $x_2 = a_{2^k+1} \cdot a_{2^k+2} \cdots a_{2^{k+1}}$. This completes the argument for the forward step.

(Backward) We will now show that if the statement is true for $k$, then it is also true for $k-1$. Assume that the statement is true for any set of $k$ positive real values, i.e. that

$\frac { x_1 + x_2 + \ldots + x_k } {k} \geq \sqrt[k]{ x_1 \times x_2 \times \cdots \times x_k}.$

Then, it will be true for the k variables

$x_1=a_1, x_2=a_2,\ldots x_{k-1} = a_{k-1} \text{ and } x_k = \frac {a_1 + a_2 + \ldots a_{k-1} } {k-1},$

implying

\begin{aligned} & \frac { a_1 + a_2 + \ldots +a_{k-1} + \frac {a_1 + a_2 + \ldots a_{k-1} } {k-1} } {k} \geq \sqrt[k]{ a_1 \times a_2 \times \cdots \times a_{k-1} \times \frac {a_1 + a_2 + \ldots a_{k-1} } {k-1} } \\ & \Leftrightarrow \frac {a_1 + a_2 + \ldots a_{k-1} } {k-1} \geq \sqrt[k]{ a_1 \times a_2 \times \cdots \times a_{k-1} \times \frac {a_1 + a_2 + \ldots a_{k-1} } {k-1} } \\ & \Leftrightarrow \left( \frac {a_1 + a_2 + \ldots a_{k-1} } {k-1} \right)^k \geq a_1 \times a_2 \times \cdots \times a_{k-1} \times \frac {a_1 + a_2 + \ldots a_{k-1} } {k-1} \\ & \Leftrightarrow \left( \frac {a_1 + a_2 + \ldots a_{k-1} } {k-1} \right)^{k-1} \geq a_1 \times a_2 \times \cdots \times a_{k-1}\\ & \Leftrightarrow \frac {a_1 + a_2 + \ldots a_{k-1} } {k-1} \geq \sqrt[k-1] { a_1 \times a_2 \times \cdots \times a_{k-1} } \end{aligned}

The last equation is the Arithmetic Mean-Geometric Mean Inequality for $k-1$ variables. This completes the proof of the backward step. $_\square$

Note by Calvin Lin
7 years ago

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