Question

Prove that the largest product P of N numbers whose sum is S is \[\rm{P}=\left(\frac{\rm{S}}{\rm{N}}\right)^\rm{N}\]
For clarity:
1. An example of a specific case would be (so you understand what I meant): Find the largest product P of 2 numbers whose sum is 42, which then solved using diff eq to obtain \[x_1=x_2=21\] and \[\rm P=x_1\cdot x_2=21\cdot 21=441\] and we can generalize that for N=2 \[\rm P=\left(\frac{\rm{S}}{\rm{2}}\right)^\rm{2}\]what I wanted to know is the proof of the general case for N
2. All numbers is a nonzero positive real number and N is a positive integer. \[x \in \mathbb{R}^+,\rm{N} \in \mathbb{Z}^+\]
3. Function to be maximized: \[\rm{P}=\prod^N_{i=1} x_i\] constraint: \[\rm{S}=\sum^N_{i=1} x_i\]

Answer 1

Proof for N=5
\[\text{constraint: }g(x,y,z,w,v)=x+y+z+w+v-\rm{S}=0 \\ \text{maximize: }\rm{P}=xyzwv \\f(x,y,z,w,v)=xyzwv+\lambda (x+y+z+w+v-\rm{S}) \\\frac{\partial f}{\partial x}=yzwv+\lambda=0 \\\frac{\partial f}{\partial y}=xzwv+\lambda=0 \\\frac{\partial f}{\partial z}=xywv+\lambda=0 \\\frac{\partial f}{\partial w}=xyzv+\lambda=0 \\\frac{\partial f}{\partial v}=xyzw+\lambda=0 \\\frac{\partial f}{\partial \lambda}=x+y+z+w+v-\rm{S}=0 \\ \frac{xyzwv}{x}=\frac{xyzwv}{y}=\frac{xyzwv}{z}=\frac{xyzwv}{w}=\frac{xyzwv}{v} \\ x=y=w=z=v \\S=5x \text{ and }P=x^5 \\ \rm{P}=\frac{S^5}{5^5} \] it is then obvious that all numbers is always equal for N>1 by the form of the differential partial, but I have no idea how to prove it rigorously, any help will be appreciated

Answer 2

You could use the arithmetic-geometric inequality maybe. If you have n non-negative numbers, then:\[(a_1a_2a_3\cdots a_n)^{\frac{1}{n}}\le \left(\frac{a_1+a_2+\cdots +a_n}{n}\right)\]with equality only a1 = a2 = ... = an.
Let S = a1 + a2 + ... + an. Taking both sides of this inequality to the nth power yields:\[a_1a_2\cdots a_n \le \left(\frac{S}{n}\right)^n\]Hence the product of the numbers is always less than or equal to (S/n)^n, with equality only happening when a1 = a2 = ... = an, which is what you want.

Answer 3

Augustus, I think you can use induction and a matrix equation with the partial derivatives. Does that help any?

Answer 4

Use subscripted \(x_i\) rather than \(x,y,z,. . .\). It will make things much easier. Also, invoke use of the notation \[\prod_{\substack{1 \leq i \leq n\\ i\neq j}}x_i\]

Answer 5

Yes joemath314159 just proved \[P \le \left(\frac{S}{n}\right)^n\] using the inequality, then it is trivial to show that \[x_i=x_j\] for all i and j. That's amazing insight, I never thought of using simple inequality. Like Limitless I thought of using induction (to no avail). Kudos to you