is this simple proof valid? (Making the product of three positive numbers maximum) I am an Egyptian engineering student,
I had a question in my assignment where i were asked to find three positive numbers whose sum is 100 and there multiplication is maximum
I know that it can be solved by finding the local maximum point ..., However i think i found another way to find the solution with a simpler/non-calculus approach, my approach was the following ...
Assume the numbers are x, y, z
And p(x, y, z) = xyz

Since 
  x + y + z
= y + z + x
= z + x + y = 100
And 
  xyz
= yzx
= zxy = p(x, y, z)

Then all the variables can replace the
 position of each other and
 maintain the same expression value

Therefor at p max
x = y = z 
And so on .....

I found that somehow make sense as why would a number be greater or less than the other with the other number being in the same position of the first number with the same degree and same everything
I am not sure about how should this be written, but i really wish to know how should i prove this in a mathematical rigorous way ....
 A: If I understand your logic correctly, you are saying that (in an alternate problem) if there are two constraints:

*

*$x + y = R$
and

*

*$x \times y = S$,

that because these two constraints are both symmetric about $x$ and $y$, the only solution possible is one where $x = y$.
Let $R = 8$ and $S = 12$, with $x,y$ both required to be in $\Bbb{R}$.
Then, the only solutions possible are $(x=2, y=6)$ and $(x=6,y=2)$.
In neither solution is $x = y$.
Edit
By analogy, the only difference between my alternate problem and your initial problem is that $S$ is supposed to represent the maximum product of $(x \times y)$ achievable, when $x + y = R.$
I see no relevant difference between that problem and my alternate problem.  If you look at all of the possible values achievable by $(x \times y)$, when $(x + y = R)$ is satisfied, either there is a distinct maximum value $S$, or there isn't.  If there is such a value, then I see no reason that $x,y$ then have to be equal, using only the logic that you have presented.
It is true that if $S$ is presumed to be the max product achievable, that then, one can (for example) invoke the comparison of geometric and arithmetic means (or the Calculus approach that you alluded to), but that is an entirely different analysis.
A: Your argument is invalid.
"Then all the variables can replace the
position of each other and
maintain the same expression value"
It's true that in a symmetric equation in $x$ and $y$ we can replace $x$ by $y$ and $y$ by $x$ and maintain the same value. But that does not mean $x$ must equal $y$. You can see that by a help of a counter example ($x+y=5$) or by noticing that we can't replace any $x$ with $y$ arbitrarily, for every $x$ we replace by $y$ we have to replace the $y$ by $x$ also. So they can't replace each other in any manner we wish as if they are equal.
Now, for a simple argument without involving calculus to find the maximum:
We have $x+y+z=100$, and we would like to achieve the maximum of the product $xyz$.
Suppse the point $(x_0,y_0,z_0)$ achieves both of these conditions.
WITHOUT LOST OF GENERALITY WE CAN ASSUME $x_0\le y_0\le z_0$.
Then the value of the product of any small deviation from this point has to be smaller or equal to the maximum value at $(x_0,y_0,z_0)$.
Now, consider $e>0$ a small number.
Then, for every $e>0$ we must have that $(x_0+e) (y_0-e) (z_0)\le x_0y_0z_0$. Notice both points do satisfy $x+y+z=100$, therefore this inequality  must hold.
With simple algebraic manipulation we arrive at:
$(y_0-x_0)\le e$ for all $e>0$ (we assumed that $0\lt z_0$)
But there is no positive number that is smaller or equal than all positive numbers ( that would imply it's smaller or equal than its half), therefore $(y_0-x_0)=0$. Similarly for $z_0$. Hence the claim.
