Proving a formula

• Feb 28th 2010, 02:15 AM
Proving a formula
Hello all,

I need help proving the following:
if a,b,c,... are n real positive number such that a+b+c+... = 1, then
a^2 + b^2 + c^2 + ... >=1/n

Thank you
• Feb 28th 2010, 02:48 AM
Bacterius
Hello,
denote all these numbers $a_1$, $a_2$, ... $a_n$, with the property :

$\sum_{k = 1}^{n} a_k = 1$

Now note that when $a_k = \frac{1}{n}$, we have $a_k^2 = \frac{1}{n^2}$, and it follows that :

$\sum_{k = 1}^{n} a_k^2 = \sum_{k = 1}^{n} \frac{1}{n^2} = n \times \frac{1}{n^2} = \frac{1}{n}$

This is not a proof in itself : it is there to help you grasp what this problem means. All you have to prove is that there is no solution for $a_k$ where $\sum_{k = 1}^{n} a_k^2 < \frac{1}{n}$.

Here is an idea : say $a_1 = \frac{1}{x}$. Therefore, the following holds : $0 < a_2 < 1 - \frac{1}{x}$, and so on following the same logic with $a_3$, $a_4$, .... Take the squares of these inequalities, and see where you get ;) You will very likely stumble upon a difficulty, that you will probably be able to express in mathematical terms and that will lead to the solution to your problem :)
• Feb 28th 2010, 03:22 AM
Hey..
I did what we said: I bounded each number between 0 and $1-\frac{1}{x}$. When added them all, I got: $\frac{1}{x^2}<\sum _{k=1}^n a_k^2 < \frac{n}{x^2} - \frac{2(n-1)}{x} + n - 1$. But i didn't know what to do next. How can I prove from here that there is no solution for the inequality $\sum_{k = 1}^{n} a_k^2 < \frac{1}{n}$?

Thank you
• Feb 28th 2010, 03:39 AM
Bacterius
EDIT : forget this stuff, I got it wrong. Hang on, I'm thinking of something ...

Yeah, I think I got it. Say you got your numbers, all equal to each other (so they all equal $\frac{1}{n}$). If you add some value $x$ to one number, you need to report the opposite of this value onto the other numbers to balance the equality. Now which values of $x$ don't allow any balancing of the other numbers that satisfies the sum of the squares being less than $\frac{1}{n}$ ? Try to express this in mathematical terms. You should find that $x$ has no solution ... your problem follows. I might post a proof, I'm actually as stumped as you on this one :D
• Feb 28th 2010, 07:44 AM
u2_wa
Quote:

Hello all,

I need help proving the following:
if a,b,c,... are n real positive number such that a+b+c+... = 1, then
a^2 + b^2 + c^2 + ... >=1/n

Thank you

If $a+b+c=1$
Let A=average of $a+b+c$
we can say that $a^2+b^2+c^2>=A^2+A^2+A^2$
Similarly,
$a+b+c+d+.....=1, A=\frac{1}{n}
$

$a^2+b^2+c^2+....>=nA^2$
$a^2+b^2+c^2+....>=\frac{1}{n}$
• Feb 28th 2010, 09:16 AM
hello u2_wa

There is one small thing I didn't get in your proof:
by saying $a^2+b^2+c^2>=A^2+A^2+A^2$, you're assuming that: $a^2>=A^2$ and $b^2>=A^2$ and $c^2>=A^2$. But if A is the average of a,b and c, then A may be smaller or larger than a, b or c. Would you please explain how did you get the inequality?

Thank you
• Mar 1st 2010, 04:15 AM
u2_wa
Quote:

hello u2_wa

There is one small thing I didn't get in your proof:
by saying $a^2+b^2+c^2>=A^2+A^2+A^2$, you're assuming that: $a^2>=A^2$ and $b^2>=A^2$ and $c^2>=A^2$. But if A is the average of a,b and c, then A may be smaller or larger than a, b or c. Would you please explain how did you get the inequality?

Thank you

Certainly I am not assuming $a^2>=A^2$ and $b^2>=A^2$ and $c^2>=A^2$

I am saying that sum of $a^2+b^2+c^2$ will always be equal to or greater than
$A^2+A^2+A^2$

To prove $a^2+b^2+c^2>=A^2+A^2+A^2$

$a^2+b^2+c^2>=3A^2$

$a^2+b^2+c^2>=3(\frac{a+b+c}{3})^2$

$\frac{3a^2+3b^2+3c^2-(a+b+c)^2}{3}>=0$

$\frac{(a-c)^2+(a-b)^2+(b-c)^2}{3}>=0$ (PROOF)

Same rule applies to $a+b+c+d+.....=nA$

Hope this helps!