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Say we have $\sum_{i=1}^n x_n = C$, i.e., $x_1+x_2+...x_n=C$ where C is a constant and $x_1,x_2,...x_n$ are nonnegative. Prove the product $(x_1)(x_2)(x_3)...(x_n)$ has a maximum if and only if $x_1=x_2=x_3=...=x_n= \frac{c}{n}$.

I’ve tried plugging in stuff like $x_1=C-x_2-x_3-x_4...-x_n$ into the product for each $x$ value to take the gradient to see if $\frac{c}{n}$ is a critical point but it ends up a really messy derivative that I don’t even know where to start on.

All help is appreciated, thank you.

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Hint: Use that $$\frac{x_1+x_2+x_3+...+x_n}{n}\geq \sqrt[n]{x_1\cdot x_2\cdot x_3\cdots x_n}$$

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  • $\begingroup$ Do you come about this by setting up $$x_1*x_2*x_3*...*x_n\le(\frac{c}{n})^n$$ and rearranging by raising both sides to 1/n, and swapping out C with $x_1+x_2+...x_n$? I see where you’re going but I don’t know if this’ll work, since it relies on the assumption that $x_1=x_2=...=x_n=\frac{c}{n}$ yields the maximum for the product, which is what I’m trying to prove. $\endgroup$ – Bob Jackson Dec 5 '18 at 8:14
  • $\begingroup$ This is the AM-GM inequality, and it is known that the equal sign holds for $$x_1=x_2=…=x_n$$ $\endgroup$ – Dr. Sonnhard Graubner Dec 5 '18 at 8:19
  • $\begingroup$ Ohh I see what you mean with the AM-GM inequality, I haven’t gone over it so I didn’t recognize it. Thanks $\endgroup$ – Bob Jackson Dec 5 '18 at 8:25
  • $\begingroup$ So your problem is solved now? $\endgroup$ – Dr. Sonnhard Graubner Dec 5 '18 at 8:26
  • $\begingroup$ I don’t think my class will accept this answer since it’s using material that hasn’t been covered, but I appreciate the really elegant method. I’ll look to see if there’s a way to do this using the AM-GM inequality without really explicitly stating it, but most likely I’ll have the use the material from last week, which is extrema and stuff with gradients and hessians of functions, and Sylvester’s criterion to find extrema. Again though, thanks. $\endgroup$ – Bob Jackson Dec 5 '18 at 8:33

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