1. ## 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

2. Hello,
denote all these numbers $\displaystyle a_1$, $\displaystyle a_2$, ... $\displaystyle a_n$, with the property :

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

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

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

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 $\displaystyle a_k$ where $\displaystyle \sum_{k = 1}^{n} a_k^2 < \frac{1}{n}$.

Here is an idea : say $\displaystyle a_1 = \frac{1}{x}$. Therefore, the following holds : $\displaystyle 0 < a_2 < 1 - \frac{1}{x}$, and so on following the same logic with $\displaystyle a_3$, $\displaystyle a_4$, .... 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

3. Hey..
I did what we said: I bounded each number between 0 and $\displaystyle 1-\frac{1}{x}$. When added them all, I got: $\displaystyle \frac{1}{x^2}<\sum _{k=1}^n a_k^2 < \frac{n}{x^2} - \frac{2(n-1)}{x} + n - 1$. But i didn't know what to do next. How can I prove from here that there is no solution for the inequality $\displaystyle \sum_{k = 1}^{n} a_k^2 < \frac{1}{n}$?

Thank you

4. 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 $\displaystyle \frac{1}{n}$). If you add some value $\displaystyle x$ to one number, you need to report the opposite of this value onto the other numbers to balance the equality. Now which values of $\displaystyle x$ don't allow any balancing of the other numbers that satisfies the sum of the squares being less than $\displaystyle \frac{1}{n}$ ? Try to express this in mathematical terms. You should find that $\displaystyle x$ has no solution ... your problem follows. I might post a proof, I'm actually as stumped as you on this one

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

$\displaystyle a^2+b^2+c^2+....>=nA^2$
$\displaystyle a^2+b^2+c^2+....>=\frac{1}{n}$

6. hello u2_wa

There is one small thing I didn't get in your proof:
by saying $\displaystyle a^2+b^2+c^2>=A^2+A^2+A^2$, you're assuming that: $\displaystyle a^2>=A^2$ and $\displaystyle b^2>=A^2$ and $\displaystyle c^2>=A^2$. 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

hello u2_wa

There is one small thing I didn't get in your proof:
by saying $\displaystyle a^2+b^2+c^2>=A^2+A^2+A^2$, you're assuming that: $\displaystyle a^2>=A^2$ and $\displaystyle b^2>=A^2$ and $\displaystyle c^2>=A^2$. 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 $\displaystyle a^2>=A^2$ and $\displaystyle b^2>=A^2$ and $\displaystyle c^2>=A^2$

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

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

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

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

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

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

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

Hope this helps!