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March 24th, 2018, 09:58 AM  #1 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,344 Thanks: 2466 Math Focus: Mainly analysis and algebra  God's Book of Proofs 
March 24th, 2018, 11:41 AM  #2 
Senior Member Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27 
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: 371 Thanks: 26 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: 6,560 Thanks: 605 
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.

March 25th, 2018, 05:14 AM  #5 
Senior Member Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27 
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  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 211 Thanks: 64 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
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*}$  

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