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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

cdm2003
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  • But you can get the left-hand side from the right? If so, do you agree the equality is legit? – Hamed Jul 04 '17 at 18:21
  • I assume the equality is legit and yes, I could probably get the left from the right. However, the text asks you to prove the inequality.alone. I had to look in the back of the book to find the "hint" which was the equality. – cdm2003 Jul 04 '17 at 18:29

3 Answers3

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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

user1952500
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Dr. Sonnhard Graubner
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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 $(r + s + t)\frac{1}{2}[(r-s)^2+(s-t)^2+(t-r)^2] \geq 0$, $r^3 + s^3 + t^3 - 3rst \geq 0$.

dcxt
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$$r^3+s^3+t^3-3trs=r^3+3r^2s+3rs^2+s^3+t^3-3r^2s-3rs^2-3rts=$$ $$=(r+s)^3+t^3-3rs(r+s+t)=(r+s+t)((r+s)^2-(r+s)t+t^2)-3rs(r+s+t)=$$ $$(r+s+t)(r^2+2rs+s^2-rt-st-3ts)=$$ $$=(r+s+t)(r^2+s^2+t^2-rs-rt-st)=\frac{1}{2}(r+s+t)((r-s)^2+(r-t)^2+(s-t)^2).$$

Michael Rozenberg
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