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Prove that $ \frac{{n!}} {{n^{\frac{n} {2}} }} $ is the max of the function $ f\left( x \right) = \prod\limits_{i = 1}^n {x_i } $ under the restriction $ g\left( x \right) = \sum\limits_{i = 1}^n {\frac{{x_i ^2 }} {{i^2 }} - 1 = 0} $.

Well Using LaGrange multiplier I have the system $ \nabla f = \lambda \nabla g $ then for each $k$ we have that $ \prod\limits_{i \ne k} {x_i } = \frac{{2\lambda x_k }} {{k^2 }} $ Thus $ \prod\limits_{i = 1}^n {x_i } = \frac{{2\lambda x_k ^2 }} {{k^2 }} $

Considering the following Sum, we conclude that:$$ n\left( {\prod\limits_{i = 1}^n {x_i } } \right) = \sum\limits_{k = 1}^n {\left( {\prod\limits_{i = 1}^n {x_i } } \right)} = \sum\limits_{k = 1}^n {\frac{{2\lambda x_k ^2 }} {{k^2 }}} = \lambda $$ I don't know how to continue :S

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Put $x_i^2=\frac{i^2}{2\lambda}\prod_{j}x_j$ in the constraint; you should obtain $1=\sum_{i=1}^n \frac{x^2_i}{i^2}=\sum_{i=1}^n \frac{\prod_{j}x_j}{2\lambda},$ i.e. $\lambda=\frac{n}{2}\prod_{j}x_j$. Put the value of the Lagrange multiplier back in $x_i^2$ and try to compute the max of $f$. – Avitus Jul 9 '13 at 8:28
I add some details and move the discussion to an answer – Avitus Jul 9 '13 at 8:34
up vote 1 down vote accepted

Put $x_i^2=\frac{i^2}{2\lambda}\prod_{j}x_j$ in the constraint; you should obtain $1=\sum_{i=1}^n \frac{x^2_i}{i^2}=\sum_{i=1}^n \frac{\prod_{j}x_j}{2\lambda},$ i.e. $$\lambda=\frac{n}{2}\prod_{j}x_j.$$

Substitute the value of the Lagrange multiplier $\lambda$ back in $x_i^2$ getting


for all $i=1,\dots,n$. Selecting the positive root to deduce the $x_i$'s, the maximum of $f$ is computed at


and is equal to

$$f(x)=\frac{1}{\sqrt{n}}\cdots\frac{n}{\sqrt{n}}=\frac{1\cdots n}{(\sqrt{n})^n}=\frac{n!}{n^{\frac{n}{2}}}.$$

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Unless this is specifically an exercise in Lagrange multipliers, it is much simpler to apply the am/gm inequality. This also has the advantage of immediately yielding that this is an absolute maximum. This is sometimes a problem with Lagrange multipliers, even to show we have a relative maximum.

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