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 n (specifically for all nn powers of 22). In the backward argument, we will show that if the statement is true for nn variables, then it is also true for n1 n-1 variables. By combining these arguments, we show that the statement is true for any value of n n, by first applying the forward argument to show that it is true for a power of 2 that is larger than nn, and then using the backward argument to show that it is true for n n.

Arithmetic Mean - Geometric Mean Inequality

AM-GM inequality: For n n non-negative real values x1,x2,xn x_1, x_2, \ldots x_n ,

i=1nxini=1nxin. \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 21=2 2^1=2 is proved in Complete the Square. For the induction step, suppose the statement is true for some 2k 2^k; we would like to show that the statement is true for 2k+1 2^{k+1}. Given {ai}i=12k+1 \{a_i\}_{i=1} ^{2^{k+1}} positive real values, we divide the set in half (obtaining {ai}i=12k \{a_i\}_{i=1} ^{2^{k}} and {ai}i=2k+12k+1 \{a_i\}_{i=2^k+1} ^{2^{k+1}}), and then apply the induction hypothesis to each set.

a1+a2++a2k+12k+1=12(a1+a2+a2k2k+a2k+1+a2k+2+a2k+12k)12(a1a2a2k2k+a2k+1a2k+2a2k+12k)a1a2a2ka2k+1a2k+2a2k+12k+1 \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 x1=a1a2a2k x_1 = a_1 \cdot a_2 \cdots a_{2^k} and x2=a2k+1a2k+2a2k+1 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 k, then it is also true for k1 k-1. Assume that the statement is true for any set of k k positive real values, i.e. that

x1+x2++xkkx1×x2××xkk. \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

x1=a1,x2=a2,xk1=ak1 and xk=a1+a2+ak1k1, 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

a1+a2++ak1+a1+a2+ak1k1ka1×a2××ak1×a1+a2+ak1k1ka1+a2+ak1k1a1×a2××ak1×a1+a2+ak1k1k(a1+a2+ak1k1)ka1×a2××ak1×a1+a2+ak1k1(a1+a2+ak1k1)k1a1×a2××ak1a1+a2+ak1k1a1×a2××ak1k1 \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 k1 k-1 variables. This completes the proof of the backward step. _\square

Note by Calvin Lin
5 years, 8 months ago

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