Problem of the Week

Problem 7 solution

(i) Let $x_1, x_2$ be a $2$ -list.

$\begin{aligned} (\sqrt{x_1}-\sqrt{x_2})^2 &\ge 0 \\ x_1 - 2\sqrt{x_1x_2} + x_2 &\ge 0 \\ x_1 + x_2 &\ge 2\sqrt{x_1x_2} \\ \underbrace{(x_1 + x_2)/2}_\alpha &\ge \underbrace{(x_1x_2)^{1/2}}_\gamma \end{aligned}$

(ii) The $kl$ -list $x_1, ..., x_kl$ can be broken down into the $k$ $l$ -lists
$(x_1, ..., x_l), ..., (x_{(k-1)l + 1}, ..., x_{kl})$ .

$\begin{aligned} \alpha &= (x_1 + ... + x_{kl})/(kl) \\ &= ( (x_1 + ... + x_l)/l + ... + (x_{(k-1)l + 1} + ... + x_{kl})/l )/k \\ &\ge ( (x_1 ... x_l)^{1/l} + ... + (x_{(k-1)l + 1} ... x_{kl})^{1/l} )/k \text{ since $\alpha \ge \gamma$ for all $l$-lists by assumption} \\ &\ge ( (x_1 ... x_l)^{1/l} ... (x_{(k-1)l + 1} ... x_{kl})^{1/l} )^{1/k} \text{ since $\alpha \ge \gamma$ for all $k$-lists by assumption} \\ &= (x_1 ... x_kl)^{1/(kl)} = \gamma \end{aligned}$

(iii)

$\begin{aligned} \alpha &= (x_1+...+x_k)/k \\ k\alpha &= x_1+...+x_k \\ (k+1)\alpha &= x_1+...+x_k+\alpha \\ \alpha &= (x_1+...+x_k+\alpha)/(k+1) \\ &\ge (x_1 ... x_k \alpha)^{1/(k+1)} \text{ since $\alpha \ge \gamma$ for all sequences of length $k+1$} \\ \alpha^{k+1} &\ge x_1 ... x_k \alpha \\ \alpha^k &\ge x_1 ... x_k \\ \alpha &\ge (x_1 ... x_k)^{1/k} = \gamma \end{aligned}$

(iv) By starting with (i) and repeatly applying (ii), $\alpha \ge \gamma$ for all $2^k$ -lists, where $k$ is a positive integer. Since there is a power of two greater than every integer, $\alpha \ge \gamma$ for all lists by repeated application of (iii).

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Problem of the Week

Problem 7 solution

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