# What do I do next

• Jun 24th 2006, 11:28 AM
Nichelle14
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
malaygoel
Quote:

Originally Posted by Nichelle14
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?

I am using a different method here.
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
malaygoel
Quote:

Originally Posted by Nichelle14
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?

I think you obtained the equation
$\displaystyle 2(a^2+b^2+c^2+ab+bc+ca)\geq0$
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 PM
Nichelle14
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 AM
rgep
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).