Factoring within a proof In the proof text I am using, I am trying to understand a proof of the fact that the geometric mean is less than or equal to the arithmetic mean by showing that:
rst $\le$ (r$^3$ + s$^3$ + t$^3$)/3
The answer in the back says to note that:
r$^3$ + s$^3$ + t$^3$ - 3rst = $\frac 12$(r + s + t)[(r - s)$^2$ + (s - t)$^2$ + (t - r)$^2$]
That said, I have no idea how they got the right side from the left, let alone how to continue with the proof. Does anyone have any pointers as to how to begin factoring the left to get the right?
Thanks!
Chris
 A: use that $$a^3+b^3+c^3-3abc=\left( a+b+c \right)  \left( {a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ac
 \right) 
$$
to prove this you can multiply it out
A: Unfortunately, I believe $r^3 + s^3 + t^3 - 3rst = (r + s + t)(r^2+s^2+t^2-rs-st-rt)$ is just an equality that is easiest to memorize. Notice that the $r^2+s^2-rs$ contains many terms found in the expansion of $(r-s)^2$. Likewise with $(s-t)^2$ and $(t-r)^2$. 
Adding these three together gives $(2r^2+2s^2+2t^2-2rs-2st-2rt)$. From here, it's not difficult to see that $(r^2+s^2+t^2-rs-st-rt) = \frac{1}{2}[(r-s)^2+(s-t)^2+(t-r)^2]$. 
Therefore, $r^3 + s^3 + t^3 - 3rst = (r + s + t)\frac{1}{2}[(r-s)^2+(s-t)^2+(t-r)^2]$.
In order to prove the inequality, you must prove that $r^3 + s^3 + t^3 - 3rst \geq 0$. Note that the AM-GM inequality you are trying to prove only works when the numbers involved are non-negative. You know that all squared numbers are non-negative, and $r+s+t$ is as well, so $(r + s + t)\frac{1}{2}[(r-s)^2+(s-t)^2+(t-r)^2]$ is also non-negative.
Because $r^3 + s^3 + t^3 - 3rst = (r + s + t)\frac{1}{2}[(r-s)^2+(s-t)^2+(t-r)^2]$, and $(r + s + t)\frac{1}{2}[(r-s)^2+(s-t)^2+(t-r)^2] \geq 0$, $r^3 + s^3 + t^3 - 3rst \geq 0$.
A: $$r^3+s^3+t^3-3trs=r^3+3r^2s+3rs^2+s^3+t^3-3r^2s-3rs^2-3rts=$$
$$=(r+s)^3+t^3-3rs(r+s+t)=(r+s+t)((r+s)^2-(r+s)t+t^2)-3rs(r+s+t)=$$
$$(r+s+t)(r^2+2rs+s^2-rt-st-3ts)=$$
$$=(r+s+t)(r^2+s^2+t^2-rs-rt-st)=\frac{1}{2}(r+s+t)((r-s)^2+(r-t)^2+(s-t)^2).$$
