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The proof goes as follows:

If $a_1, \dots, a_n$ are positive numbers then: $$ (a_1 + \dots + a_n)/n \geq \sqrt[n]{a_1 \dots a_n} $$ The left-hand side we denote by $A_n$ and the right-hand side by $G_n$.

The inequality is trivial when $n = 1$. Assume the inequality is true if $n = k$.

Applying this inequality to $b_1 = a_{k+1}$ and $b_2 = b_3 = b_k = A_{k+1}$, we get: $$ \frac{a_{k+1} + (k-1)A_{k+1}}{k} \geq (a_{k+1}A_{k+1}^{k-1})^{1/k} $$ Now let: $$ A = \frac{a_{k+1} + (k-1)A_{k+1}}{k} \quad and \quad G = (a_{k+1}A_{k+1}^{k-1})^{1/k} \tag{1} $$

Applying the inequality for the case in which n = 2, we get: $$ A_{k+1} = \frac{A_k + A}{2} \geq (A_k A)^{1/2} \geq (G_k G)^{1/2} = (G_{k+1}^{k+1}A_{k+1}^{k-1})^{1/2k} \tag{2} $$ That is $$A_{k+1} \geq (G_{k+1}^{k+1}A_{k+1}^{k-1})^{1/2k}$$ and thus $A_{k+1} \geq G_{k+1}$.

Specifically once we define $A$ and $G$ in (1) I fail to get the meaning of the subscripts for $A$ and $G$ in (2). I mainly tried to solve my difficulty using $G_k$, but every time I cannot unfold $G_k G$ to get $(G_{k+1}^{k+1}A_{k+1}^{k-1})^{1/2k}$.

limah
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1 Answers1

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Notice that we earlier defined$$A_k=\frac{a_1+a_2+\cdot\cdot\cdot+a_k}k\\G_k=(a_1a_2...a_k)^{1/k}$$Thus$$\frac{A_k+A}2=\frac{\frac{a_1+a_2+\cdot\cdot\cdot+a_k}k+\frac{a_{k+1}+(k-1)A_{k+1}}k}2=\frac{(k+1)A_{k+1}+(k-1)A_{k+1}}{2k}=A_{k+1}$$The inductive step gives $A_k\ge G_k$ and $A\ge G$ and thus $(A_kA)^{1/2}\ge(G_kG)^{1/2}$.

Further, $G_kG=(a_1a_2...a_k)^{1/k}\left(a_{k+1}A^{k-1}_{k+1}\right)^{1/k}=\left[\underbrace{(a_1a_2...a_{k+1})}_{G_{k+1}^{k+1}}\left(A_{k+1}^{k-1}\right)\right]^{1/k}$ from where we get our result.

Shubham Johri
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