<s>[INST] Showing $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$ for $x,y,z \ge 0$ How do I show that $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$ for $x,y,z \ge 0$? 
The answer is to: 

Let $A = x+y+z$, then $z=A-x-y$. Maximize $f(x,y)=xy(A-x-y)$ on region enclosed by $x=y=0$ and $x+y=A$

What I don't understand is the idea behind this method. 


*

*Why $A=x+y+z$ not $A=\frac{x+y+z}{3}$? 

*I guess $f(x,y)=xy(A-x-y)$ is to see whats the maximum of the equation

*But why regions $x=y=0$ and $x+y=A$? 



Sorry this seems like an easy question but somehow I don't get it ... 
 [/INST] For your first question, the answer is that it doesn't matter. We could just as well let $A=\frac{x+y+z}{3}$. Then $z=3A-x-y$, and so on. The calculations end up very similar. Indeed things are a little prettier, I think, if we let $\frac{x+y+z}{3}=A$.  But it makes no real difference.
The point is that we fix $x+y+z$, and show using tools from the calculus that for the fixed value $x+y+z=A$, we must have $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$.  Since our argument works for any non-negative $A$, that shows that the desired inequality always holds.
In order to show that $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$, we find the maximum value of $xyz$, given that $x+y+z=A$ and $x,y,z\ge 0$.  Since $z=A-x-y$, we want to find the maximum value of $f(x,y)=xy(A-x-y)$. It will turn out when you maximize that this maximum value is $A^3/27$, which gives exactly the right result.  For it implies that the maximum of $\sqrt[3]{xyz}$ given that $x+y+z=A$ is $\sqrt[3]{A^3/27}$, which is $A/3$.
Of course we will only consider $x,y\ge 0$. The condition $z\ge 0$ becomes, in terms of $x$ and $y$, that $A-x-y \ge 0$. This can be rewritten as $x+y \le A$.
So we are maximizing the function (not equation) $xy(A-x-y)$ in the region that satisfies $x\ge 0$, $y\ge 0$, $x+y \le A$.
This is the same problem as maximizing $xy(A-x-y)$ over the triangle bounded by the lines $x=0$, $y=0$, and $x+y=A$. (The condition $x+y \le A$ just says that $(x,y)$ lies below the line $x+y=A$.)
I am sure you can handle the partial derivatives stuff.
Remark:  Actually, because the inequality we are interested in is homogeneous (it holds at $(x,y,z)$ iff it holds at $(tx,ty,tz)$), we could even assume that $x+y+z$ is some specific number, like $3$.
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