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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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Quote:
 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.

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