1. $\prod_{i=1}^n x_i = 1$ leads to constraint function $G(x_1,x_2,...,x_n)=\prod_{i=1}^n x_i-1$

($\prod_{i=1}^n x_i =x_1 x_2...x_n$)

Task is to to find the minimum using conditional extrema of the following (the induction method that is most convinient is forbidden), if we proove this special case then the derivation can be generalised to prove the AMGM theorem:

  1. $F(x_1,x_2,...,x_n) = \sum_{i=1}^n x_i$

($\sum_{i=1}^n x_i =x_1+x_2+...+x_n$)

Idea is that it should be $\sum_{i=1}^n x_i \geqslant n$ afaik

And finally Using the derivation above prove the AM-GM theorem:

$\frac{\sum_{i=1}^n x_i}{n} \geqslant \sqrt[n]{\prod_{i=1}^n x_i - 1}$

Solution so far:

what I come up with is writing down the Lagrangian:

$L = F(x_1,x_2,...,x_n) - \lambda G(x_1,x_2,...,x_n) \implies L = \sum_{i=1}^n x_i - \lambda (\prod_{i=1}^n x_i-1)$

taking the partial derivatives

$\frac{dL}{dx_1} = 1 - \lambda \frac{\prod_{i=1}^n x_i}{x_1}=0 \implies \lambda \frac{\prod_{i=1}^n x_i}{x_1}=1 \implies \lambda \frac{1}{x_1}=1 \implies \lambda = x_1$

$\frac{dL}{dx_2} = 1 - \lambda \frac{\prod_{i=1}^n x_i}{x_2}=0 \implies \lambda \frac{\prod_{i=1}^n x_i}{x_2}=1 \implies \lambda \frac{1}{x_2}=1\implies \lambda = x_2$


$\frac{dL}{dx_n} = 1 - \lambda \frac{\prod_{i=1}^n x_i}{x_n}=0 \implies \lambda \frac{\prod_{i=1}^n x_i}{x_n}=1 \implies \lambda \frac{1}{x_n}=1 \implies \lambda = x_n$

$\frac{dL}{d\lambda} = - \prod_{i=1}^n x_i + 1= 0 \implies$ $\prod_{i=1}^n x_i = 1$

$ \lambda = x_1 = x_2 = ...=x_n = 1$ is our critical point.

Taking the differential of our constraint

$dG(x_1,x_2,...,x_n) = 0$

$\frac{\partial G}{\partial x_1}\Delta x_1+\frac{\partial G}{\partial x_2}\Delta x_2+...+\frac{\partial G}{\partial x_n}\Delta x_n=0$

$\frac{\prod_{i=1}^n x_i}{x_1}\Delta x_1 + \frac{\prod_{i=1}^n x_i}{x_2}\Delta x_2+...+\frac{\prod_{i=1}^n x_i}{x_n}\Delta x_n=0$

substituting the roots of critical point $x_1=x_2=...=x_n=1$ and $\prod_{i=1}^n x_i=1$ leads to

$\Delta x_1 + \Delta x_2 +...+ \Delta x_n = 0$

First question.

update, answered by Andreas below

The second order differential of Lagrangian has to be positive, but I’m getting negative sign

$d^2L = \sum_{j=1}^n\sum_{i=1}^n L_{x_j x_i} \Delta x_j \Delta x_i=-\lambda (\prod x_i )(\frac{\Delta x_i \Delta x_j}{x_i x_j})<0$

Here I took the second order partial derivatives of L

$\frac{d^2 L} {dx_1 dx_1} = -\lambda \frac{\prod x_i}{x_1 x_1}$

$\frac{d^2 L} {dx_1 dx_2} = -\lambda \frac{\prod x_i}{x_1 x_2}$


$\frac{d^2 L} {dx_n dx_n} = -\lambda \frac{\prod x_i}{x_n x_n}$

Second question. Are these reasonings correct?

It is needed to justify why local extrema is global as well.

If the second differential will become positive and therefore at the point (1,1,...,1) is local minima(since the function might be just like a cubic polynomial with no global minima), then we can check the function limit along axis direction (by sending all but one variable to infinity and getting the that one variable limit to zero) and thus we find out that the function F is located in the positive n-dimensional quadrant and therefore the extreme point has to be global minima. I don't know it's just a feeling, was thinking of rotating the function 45 degrees towards the vertical axis and then stating that function goes to infinity in each direction.

Update. Probs solved the global issue part, gonna consilt with Professor and update the solution if my assumptions are right.

Update. Just posted Proof to Wiki: MyProof Idea was to simply use the Weierstrass theorem and apply it to any closed domain interval of function.


1 Answer 1


Taking the second derivatives and evaluating at $\lambda = x_1 = x_2 = ...=x_n = 1$ gives $$ L = \sum_{i=1}^n x_i - \lambda (\prod_{i=1}^n x_i-1)\\ \frac{dL}{dx_k} = 1 - \lambda \frac{\prod_{i=1}^n x_i}{x_k}\\ \frac{d^2L}{dx_k^2} = 0\\ \frac{d^2L}{dx_k d x_m} = - \lambda \frac{\prod_{i=1}^n x_i}{x_k x_m} = -1 (k \ne m) $$ Now you need that for any vector $\Delta x$ with $\Delta x_1 + \Delta x_2 +...+ \Delta x_n = 0$ holds: $$\sum_{i=1}^n\sum_{j=1}^n \frac{d^2L}{dx_i d x_j} \Delta x_i \Delta x_j > 0$$. We have $$\sum_{i=1}^n\sum_{j=1}^n \frac{d^2L}{dx_i d x_j} \Delta x_i \Delta x_j = - \sum_{i=1}^n\sum_{j \neq i} \Delta x_i \Delta x_j \\ = - \sum_{i=1}^n \Delta x_i \sum_{j \neq i} \Delta x_j \\ = - \sum_{i=1}^n \Delta x_i (0 -\Delta x_i ) = \sum_{i=1}^n \Delta x_i^2 > 0$$.

  • $\begingroup$ Thanks! Figured out as well. Now left with the local to global part to think of. $\endgroup$
    – user119510
    Nov 21, 2018 at 14:50

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