Problem 9 M.L Krasnov variational calculus Warning: Finding extreme value of a multivariable function My question differs from this since I try to use the Hessian criterion so it is not a repeated question.
My question: Problem 9 M.L Krasnov.
Find extrema of $f:U \to \mathbb{R}$ where $U=\{(x_1,\ldots,x_n)\in \mathbb{R}^n:x_{i}>0 \text{ for all }i\in \mathbb{N}\}$ and
$$f(x_1,x_2,\ldots,x_n)=x_1x_2^2\ldots x_n^n(1-x_1-2x_2-\ldots-nx_n). $$
Well this is my attempt:
Critical points are:
$$\frac{\partial f}{\partial x_n}=nx_1x_2^2x_3^3(1-x_1-2x_2-\cdots-nx_n)-nx_1x_2^2\cdots x_n^n=0$$
so $$x_1=x_2=\cdots =x_n=\frac{2}{n^2+n+2}.$$
These are critical points of $f$.
Computing second partial derivatives for use Hessian in critical point $x_0=\left(\frac{2}{n^2+n+2},\ldots, \frac{2}{n^2+n+2}\right)$ and suppose no loss of generality $i<j<n$ I get:
$$\frac{\partial^2 f}{\partial x_i^2}=i(i-1)\left(\frac{2}{n^2+n+2}\right)^{\frac{n(n+1)}{2}}-2i^2\left(\frac{2}{n^2+n+2}\right)^{\frac{n(n+1)}{2}-1}.$$
So computing the cross derivatives and substituting into $x_1=x_2=\cdots=x_n=(\frac{2}{n^2+n+2})$
$$\frac{\partial^2 f}{\partial x_j \partial x_i}=-ij\left(\frac{2}{n^2+n+2}\right)^{\frac{n(n+1)}{2}-1}.$$
I have tried this exercise for many days without any success, I plan to use Sylvester's criterion to show that this critical point is a local maximum.
Sylvester's criterion: The real-symmetric matrix $A$ is positive definite if and only if all the leading principal minors of $A$ are positive.
Help me please.
 A: *

*OP's function is
$$f(x)~:=~\pi(x)(1-\sigma(x)),\qquad 
x~=~(x_1,\ldots,x_n)~\in~ \mathbb{R}^n_{+},\tag{1}$$
$$ \pi(x)~:=~ \prod_{i=1}^n x_i^i, \qquad 
\sigma(x)~:=~\sum_{i=1}^n ix_i,\tag{2}$$
$$ \frac{\partial \pi(x)}{\partial x_j}~=~ \frac{j \pi(x)}{ x_j},\qquad \frac{\partial \sigma(x)}{\partial x_j}~=~ j,\tag{3}$$
$$ \frac{\partial f(x)}{\partial x_j} 
~=~j\pi(x)\left( \frac{1-\sigma(x)}{x_j}-1\right)
~=~j\pi(x) \frac{1-\sigma(x)-x_j}{x_j}.\tag{4}$$


*We see from eq. (4) that critical points can only be on the diagonal $$x_1~=~\ldots~=~x_n~=~1-\sigma(x).\tag{5}$$
It follows immediately that the only critical point is
$$x_1~=~\ldots~=~x_n
~=\left(1+\sum_{i=1}^n i\right)^{-1}
~=\left(1+\frac{n(n+1)}{2}\right)^{-1}
~=~\frac{2}{n^2+n+2},\tag{6}$$
as OP correctly states.


*User copper.hat gave here a simple argument why the maximum must be attained at an interior point. Hence the above critical point (6) must be a maximum. However OP wants instead to argue for a maximum via the Hessian.


*Hessian
$$\frac{\partial^2 f(x)}{\partial x_j\partial x_k}
~=~\frac{jk\pi(x)}{x_k}\left( \frac{1-\sigma(x)}{x_j}-1\right) - j\pi(x)\left(\frac{k}{x_j}+ \frac{1-\sigma(x)}{x_j^2}\delta_{jk} \right). \tag{7}$$
Hessian on-shell:
$$\left. \frac{\partial^2 f(x)}{\partial x_j\partial x_k}\right|
~=~-j(k+\delta_{jk})\frac{\pi(x)}{x_j}.\tag{8}$$


*To prove that the critical point (6) is a maximum, it is enough to prove that the matrix
$$C_{jk}~=~j(k+\delta_{jk})~=~A_{jk}+B_{jk}\tag{9}$$
is a positive matrix. The second term $$B_{jk}~=~j\delta_{jk}\tag{10}$$
is clearly a positive matrix. Since (semi)positive matrices form a convex cone, it is enough to prove that the first term
$$A_{jk}~=~jk\tag{11}$$
is a semi-positive matrix, i.e.
$$\forall v~\in~\mathbb{C}^n:~~v^{\dagger}A v~\geq~0.\tag{12}$$
By scaling $w_i=iv_i$, this is equivalent to proving that the matrix
$$D_{jk}~=~1,\tag{13}$$ or equivalently that the matrix
$$P_{jk}~=~\frac{1}{n}D_{jk}~=~\frac{1}{n},\tag{14}$$
is a semi-positive matrix. But $$P^2~=~P~=~P^{\dagger}\tag{15} $$
is an orthogonal projection matrix, and hence semi-positive. $\Box$
