An interesting inequality

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

So, to solve this problem, a substitution must be made. Let $y_i = \frac{1}{2} - x_i$ . Therefore $0 < y_i < \frac{1}{2}$ for all $i$ . Also of note, $\frac{1}{2} + x_i = 1 - y_i$ is easily verified. By this definition, $a_1 = 1 - y_1, a_2 = (1 - y_2) y_1, a_3=(1 - y_3) y_2 y_1,...$ or in general, $a_i = (1 - y_i) \displaystyle \prod_{i=1}^{i - 1} y_i = \displaystyle \prod_{i=1}^{i - 1} y_i - \displaystyle \prod_{i=1}^i y_i$ . I'll let the reader explore why the forth statement given isn't necessary. So from this, $\displaystyle \sum_{k=1}^\infty ka_k = 1 - y_1 + 2y_1 - 2y_1y_2 +3y_1y_2 - 3y_1y_2y_3 + 4y_1y_2y_3y_4 \ldots = 1 + y_1 + y_1y_2 + y_1y_2y_3 + \ldots$ . The only way to maximize the sum is to have every $y_i = \frac{1}{2}$ . Doing so produces 2, and thus equality with the final statement. All other $y_i$ are less than that. For more rigor, one can show that the previous sum is $1 + y_1(1 + y_2(1 + y_3(1 + \ldots$ , each parentheses of which must be less than 2. The answer becomes much more apparent when you make the intial substitution.

Bob Krueger - 8 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$