- #1

EnlightenedOne

- 48

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## Homework Statement

Let ## a,b,c \in \mathbb{R}^{+} ##.

Prove that $$ \sqrt[3]{abc} \leq \frac{a+b+c}{3}. $$

Note: ## a,b,c ## can be expressed as ## a = r^3, b = s^3, c = t^3 ## for ## r,s,t > 0##.

## Homework Equations

## P(a,b,c): a,b,c \in \mathbb{R}^{+} ##

## Q(a,b,c): \sqrt[3]{abc} \leq \frac{a+b+c}{3} ##

## The Attempt at a Solution

On the side, I tried to substitute ## a = r^3, b = s^3, c = t^3 ## to see where it would take me, and I ended up here:

## r^3 + s^3 + t^3 -3rst \geq 0 ##

So, in order to directly prove the original statement, I would first have to start with a statement involving ## r,s,t \geq 0 ## and then work backwards to ## Q(a,b,c) ##.

In other words, before I begin my proof I need to show that ## r^3 + s^3 + t^3 -3rst \geq 0 ##, but in order to do that I would have to factor it somehow.

The solution says:

"Observe that ## r^3 + s^3 + t^3 -3rst = \frac{1}{2} (r+s+t)[(r-s)^2 + (s-t)^2 +(t-r)^2] ##"

So, my question is:

**How would I have factored $$ r^3 + s^3 + t^3 -3rst $$ to $$ \frac{1}{2} (r+s+t)[(r-s)^2 + (s-t)^2 +(t-r)^2] $$ without using software? (paper and pencil)**It seems impossible that I would have ever been able to factor that.

Thanks!