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 March 24th, 2018, 09:58 AM #1 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra God's Book of Proofs A diverting article on an interesting book. https://www.wired.com/story/in-searc...ect-proofs/amp Thanks from topsquark and AplanisTophet March 24th, 2018, 11:41 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 603 Thanks: 88 Let's show one of them. Prove that $\displaystyle a_1 ^2 +a_2 ^2 +...+a_n^ 2 \geq \frac{1}{n} for \; \; a_1 + a_2 + ... + a_n = 1$ $\displaystyle a_1 , a_2 ,..., a_n$ real numbers Last edited by skipjack; March 24th, 2018 at 11:54 AM. March 24th, 2018, 12:17 PM #3 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 444 Thanks: 29 Math Focus: Number theory Beautifully stated, yet I have no idea how to prove it. Ramanujan? Last edited by Loren; March 24th, 2018 at 12:25 PM. March 24th, 2018, 12:40 PM #4 Global Moderator   Joined: May 2007 Posts: 6,807 Thanks: 717 Start from assuming all $a_k=\frac{1}{n}$. Then sum $=\frac{1}{n}$. Show that any deviation increases the sum. You might be able to use mathematical induction starting from n=2. Thanks from topsquark March 25th, 2018, 05:14 AM #5 Senior Member   Joined: Dec 2015 From: somewhere Posts: 603 Thanks: 88 I will show at least one method ... $\displaystyle (1^2 + 1^2 +... + 1^2)\sum a_\omega ^2 \geq \sum a_\omega$ $\displaystyle \Rightarrow n\sum a_\omega ^2 \geq ( \sum a_\omega )^2 = 1^2 = 1$ proved March 25th, 2018, 07:15 AM   #6
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 Originally Posted by idontknow $\displaystyle (1^2 + 1^2 +... + 1^2)\sum a_\omega ^2 \geq \sum a_\omega$
Perhaps I'm being a bit slow, but it's not immediately clear to me why this is true.

It's not the nicest, but here's what I came up with: for each $i$, we have $(\frac{1}{n} - a_i)^2 = \frac{1}{n} (\frac{1}{n} - 2a_i) + a_i^2$, so

\begin{align*} \sum_{i=1}^{n} {a_i^2} &= \sum_{i=1}^{n} \left(\frac{1}{n} - a_i \right)^2 + \frac{1}{n} \sum_{i=1}^{n} \left( 2a_i - \frac{1}{n} \right) \\ &\geq \frac{1}{n} \sum_{i=1}^{n} \left( 2a_i - \frac{1}{n} \right) \qquad \qquad & \text{since\left(\frac{1}{n} - a_i \right)^2 \geq 0$for all$i$} \\ &= \frac{1}{n} \left( 2\sum_{i=1}^{n} a_i - \sum_{i=1}^{n} \frac{1}{n} \right) \\ &= \frac{1}{n} \left(2 \times 1 - n \times \frac{1}{n}\right) & \text{since$\sum_{i=1}^{n}a_i = 1} \\ &= \frac{1}{n} \end{align*} January 9th, 2019, 11:05 PM   #7
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 Originally Posted by cjem Perhaps I'm being a bit slow, but it's not immediately clear to me why this is true. It's not the nicest, but here's what I came up with: for each $i$, we have $(\frac{1}{n} - a_i)^2 = \frac{1}{n} (\frac{1}{n} - 2a_i) + a_i^2$, so \begin{align*} \sum_{i=1}^{n} {a_i^2} &= \sum_{i=1}^{n} \left(\frac{1}{n} - a_i \right)^2 + \frac{1}{n} \sum_{i=1}^{n} \left( 2a_i - \frac{1}{n} \right) \\ &\geq \frac{1}{n} \sum_{i=1}^{n} \left( 2a_i - \frac{1}{n} \right) \qquad \qquad & \text{since\left(\frac{1}{n} - a_i \right)^2 \geq 0$for all$i$} \\ &= \frac{1}{n} \left( 2\sum_{i=1}^{n} a_i - \sum_{i=1}^{n} \frac{1}{n} \right) \\ &= \frac{1}{n} \left(2 \times 1 - n \times \frac{1}{n}\right) & \text{since$\sum_{i=1}^{n}a_i = 1} \\ &= \frac{1}{n} \end{align*}
How did you get this:

$\displaystyle \sum_{i=1}^{n} \frac{1}{n} = n \times \frac{1}{n}$

Isn't $\displaystyle \sum_{i=1}^{n} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$ ?

Last edited by pnerd; January 9th, 2019 at 11:08 PM. January 10th, 2019, 05:43 AM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra No $$\sum_{i=1}^{n} \frac{1}{i} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$$ January 10th, 2019, 06:36 AM   #9
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 Originally Posted by v8archie No $$\sum_{i=1}^{n} \frac{1}{i} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$$
Oh yes. I had a brain fart. January 10th, 2019, 07:19 AM #10 Senior Member   Joined: Dec 2015 From: somewhere Posts: 603 Thanks: 88 It is solved but symbols are not used well . Last edited by idontknow; January 10th, 2019 at 07:22 AM. Tags book, god, proofs Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post tsg91 Number Theory 1 February 24th, 2014 06:39 PM john898 Real Analysis 1 October 19th, 2011 04:17 AM rowdy3 Calculus 12 January 31st, 2011 07:06 PM Cushing Number Theory 3 October 17th, 2010 11:54 PM N010289 Algebra 1 October 22nd, 2007 10:24 PM

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