I am trying to prove that for any real numbers a,b,c the follwing inequality holds:

(a^2 + b^2 + c^2)/3 >= ((a+b+c)/3)^2

I multiplied out the square on the RHS and simplified.

Then got rid of the fraction on each side.

But what do I do next?

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- Jun 24th 2006, 11:28 AM #1

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## What do I do next

I am trying to prove that for any real numbers a,b,c the follwing inequality holds:

(a^2 + b^2 + c^2)/3 >= ((a+b+c)/3)^2

I multiplied out the square on the RHS and simplified.

Then got rid of the fraction on each side.

But what do I do next?

- Jun 24th 2006, 06:33 PM #2Originally Posted by
**Nichelle14**

LHS=$\displaystyle \frac{a^2 + b^2 + c^2}{3}=\frac{3(a^2 + b^2 + c^2)}{9}$

$\displaystyle =\frac{a^2 + b^2 + c^2}{9}+\frac{2(a^2 + b^2 + c^2)}{9}$

$\displaystyle \geq\frac{a^2 + b^2 + c^2}{9}+\frac{2(ab+bc+ca)}{9}$

$\displaystyle \geq\frac{(a + b + c)^2}{9}$

Keep Smiling

Malay

- Jun 24th 2006, 06:36 PM #3

- Jun 24th 2006, 06:39 PM #4

- Joined
- Jun 2006
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- Jun 25th 2006, 03:21 AM #5
If x denotes a sequence of numbers, and AM(x) its arithmetic mean, you're being asked to prove that AM(f(x)) >= f(AM(x)) where f is squaring and f(x) means the sequence of values of f applied to the elements of x. You could prove this using only the fact that f is convex (f' and f'' both positive).