Finding maximum of function of $n$ variables.

Question

Let $a,b \in \mathbb{R}$ such that $0<a<b$ and let $n \in \mathbb{N}$. Consider the function $$f(x_1,...x_n)=\frac{(x_1\cdot... \cdot x_n)}{(a+x_1)(x_1+x_2)...(x_{n-1}+x_n)(x_n+b)}$$ where $x_i \in [a,b]$ and $n$ is fixed. Find the maximum of such function.

I was trying to calculate partial derivatives and find if the maximum occurs in the interior. If we write for symmetry that $x_0=a$ and $x_{n+1}=b$ i got that $\frac{df}{dx_i}= (x_{i-1}x_{i+1} -x_i^2) \cdot S$, where S is nonzero for $i=1,2,...,n$. So maximum in the interior might occure iff $x_{i-1}x_{i+1}-x_i^2=0$ for $i=1,2,...,n$. But then I also have to check boundary.

Are partial derivatives good way to tackle this problem or should I try with some inequalities? Any help will be greatly appreciated.

Answer 1

One can use

• Show that $1+(x_1x_2...x_n)^{\frac{1}{n}} \leq [(1+x_1)(1+x_2)...(1+x_n)]^{\frac{1}{n}}$

(which is an application of Jensen's inequality to the convex function $g(t)=\log(1+e^t)$, to estimate the reciprocal of $f(x_1, \cdots, x_n)$: $$ \begin{align} \frac{1}{f(x_1, \ldots, x_n)} &= a \left( 1 + \frac{x_1}{a}\right)\left( 1 + \frac{x_2}{x_1}\right) \cdots \left( 1 + \frac{x_n}{x_{n-1}}\right)\left( 1 + \frac{b}{x_n}\right) \\ &\ge a \left( 1 + \left( \frac{x_1}{a} \frac{x_2}{x_1} \cdots \frac{x_n}{x_{n-1}}\frac{b}{x_n}\right)^{1/(n+1)}\right)^{n+1} \\ &= a \left( 1 + \left( \frac ba\right)^{1/(n+1)}\right)^{n+1} \\ &= \left( a^{1/(n+1)} + b^{1/(n+1)}\right)^{n+1} \, , \end{align} $$ so that $$ f(x_1, \ldots, x_n) \le \frac{1}{\left( a^{1/(n+1)} + b^{1/(n+1)}\right)^{n+1}} \, . $$ Equality holds if $$ \frac{x_1}{a} =\frac{x_2}{x_1} =\cdots =\frac{x_n}{x_{n-1}}=\frac{b}{x_n} \, , $$ that is if $(a, x_1, x_2, \ldots, x_n, b)$ form a geometric progression.