- #1
EnlightenedOne
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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!