=== Necessary Condition === Let: :$\forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j = x$ Then: {{begin-eqn}} {{eqn | l = A_n       | r = \dfrac 1 n \sum_{j \mathop = 1}^n x       | c =  }} {{eqn | r = \dfrac 1 n n x       | c =  }} {{eqn | r = x       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = G_n       | r = \paren {\prod_{j \mathop = 1}^n x}^{\frac 1 n}       | c =  }} {{eqn | r = \paren {x^n}^{\frac 1 n}       | c =  }} {{eqn | r = x       | c =  }} {{end-eqn}} So: :$A_n = G_n = n$ {{qed|lemma}} === Sufficient Condition === Let $A_n = G_n$. We prove the result by induction: For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :$A_n = G_n \implies \forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$ $\map P 1$ is true, as this just says: {{begin-eqn}} {{eqn | l = A_1       | r = G_1       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 1 \sum_{j \mathop = 1}^1 x_j       | r = \paren {\prod_{j \mathop = 1}^1 x_j}^{1/1}       | c =  }} {{eqn | ll= \leadsto       | l = x_1       | r = x_1       | c =  }} {{end-eqn}} which is trivially true. ==== Basis for the Induction ==== $\map P 2$ is the case: {{begin-eqn}} {{eqn | l = A_2       | r = G_2       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {x_1 + x_2} 2       | r = \sqrt {x_1 x_2}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\dfrac {x_1 + x_2} 2}^2       | r = x_1 x_2       | c =  }} {{eqn | ll= \leadsto       | l = x_1^2 + 2 x_1 x_2 + x_2^2       | r = 4 x_1 x_2       | c =  }} {{eqn | ll= \leadsto       | l = x_1^2 - 2 x_1 x_2 + x_2^2       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x_1 - x_2}^2       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x_1       | r = x_2       | c =  }} {{end-eqn}} This is our basis for the induction. ==== Induction Hypothesis ==== Now we show that: :$(1): \quad$ If $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {2 k}$ is true :$(2): \quad$ If $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k - 1}$ is true. The result will follow by Backwards Induction. This is our first induction hypothesis: :$A_k = G_k \implies \forall i, j \in \set {1, 2, \ldots, k}: x_i = x_j = x$ Also, let: :$A_k' := \displaystyle \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j = y$ and: :$G_k' := \displaystyle \paren {\prod_{j \mathop = k + 1}^{2 k} }^{1 / k} x_j = y$ By the induction hypothesis: :$A_k' = G_k' \implies \forall i, j \in \set {k + 1, k + 2, \ldots, 2 k}: x_i = x_j = y$ We need to show: :$A_{2 k} = G_{2 k} \implies \forall i, j \in \set {1, 2, \ldots, 2 k}: x_i = x_j = x$ ==== Induction Step ==== This is our induction step: Suppose: :$A_{2 k} = G_{2 k}$ Let: :$A_k' := \displaystyle \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j = y$ and: :$G_k' := \displaystyle \paren {\prod_{j \mathop = k + 1}^{2 k} }^{1 / k} x_j = y$ By the induction hypothesis: :$A_k' = G_k' \implies \forall i, j \in \set {k + 1, k + 2, \ldots, 2 k}: x_i = x_j = y$ Then: {{begin-eqn}} {{eqn | l = A_{2 k}       | r = \dfrac 1 {2 k} \sum_{j \mathop = 1}^{2 k} x_j       | c = {{Defof|Arithmetic Mean}} }} {{eqn | r = \dfrac 1 {2 k} \paren {\sum_{j \mathop = 1}^k x_j + \sum_{j \mathop = k + 1}^{2 k} x_j}       | c =  }} {{eqn | r = \dfrac 1 2 \paren {\dfrac 1 k \sum_{j \mathop = 1}^k x_j + \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j}       | c =  }} {{eqn | r = \dfrac 1 2 \paren {A_k + A_k'}       | c = {{Defof|Arithmetic Mean}} }} {{eqn | r = \dfrac 1 2 \paren {x + y}       | c = Definition of $A_k$ and $A_k'$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = G_{2 k}       | r = \paren {\prod_{j \mathop = 1}^{2 k} x_j}^{1 / 2 k}       | c = {{Defof|Geometric Mean}} }} {{eqn | r = \paren {\paren {\prod_{j \mathop = 1}^{2 k} x_j}^{1 / k} }^{1 / 2}       | c =  }} {{eqn | r = \paren {\paren {\prod_{j \mathop = 1}^k x_j}^{1 / k} \times \paren {\prod_{j \mathop = k + 1}^{2 k} x_j}^{1 / k} }^{1 / 2}       | c =  }} {{eqn | r = \paren {G_k G_k'}^{1 / 2}       | c = {{Defof|Geometric Mean}} }} {{eqn | r = \paren {x y}^{1 / 2}       | c = Definition of $G_k$ and $G_k'$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = A_{2 k}       | r = G_{2 k}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 2 \paren {x + y}       | r = \paren {x y}^{1 / 2}       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = y       | c = Basis for the Induction }} {{end-eqn}} That is: :$\forall i, j \in \set {1, 2, \ldots, k}: x_i = x_j = x$ :$\forall i, j \in \set {k + 1, k + 2, \ldots, 2 k}: x_i = x_j = x$ Hence: :$\forall i, j \in \set {1, 2, \ldots, 2 k}: x_i = x_j = x$ Now suppose $\map P k$ holds. Then: {{begin-eqn}} {{eqn | l = G_{k - 1}       | r = A_{k - 1}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {k - 1} G_{k - 1}       | r = \paren {k - 1} A_{k - 1}       | c =  }} {{eqn | ll= \leadsto       | l = k G_{k - 1}       | r = \paren {k - 1} A_{k - 1} + G_{k - 1}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {G_{k - 1}^{k - 1} G_{k - 1} }^{1/k}       | r = \dfrac {\paren {k - 1} A_{k - 1} + G_{k - 1} } k       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}       | r = \dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}       | r = \dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k       | c =  }} {{end-eqn}} But: :$\paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}$ is the geometric mean of $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$ :$\dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k$ is the arithmetic mean of $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$ We have that $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$ has $k$ elements. Hence by the induction hypothesis: :$\forall i, j \in \set {1, 2, \ldots, k - 1}: x_i = x_j = G_{k - 1}$ So $\map P k \implies \map P {k - 1}$ and the result follows by Backwards Induction. Therefore $A_n \ge G_n$ for all $n$. :$\forall n \in \N: A_n = G_n \implies \forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$ {{qed}}