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 AMNichelle14What 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 PMmalaygoelQuote:

Originally 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 PMmalaygoelQuote:

Originally Posted by**Nichelle14**

$\displaystyle 2(a^2+b^2+c^2+ab+bc+ca)\geq0$

Do like this

$\displaystyle (a^2+b^2+2ab)+(b^2+c^2+2bc)+(c^2+a^2+2ac)\geq0$

KeepSmiling

Malay - Jun 24th 2006, 06:39 PMNichelle14
that's clever.

I did get down to

a^2 + b^2 +c^2 >= ab + bc + ac

and figured it out. I like your way too! :) - Jun 25th 2006, 03:21 AMrgep
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).