A general question on comparison of integrals and a specific problem

Question

When working on an applied math topic, I have come across the following general problem.

Let $f(x_1, x_2, ..., x_n)$ be a real function of $n$ real variables $x_1, x_2, ..., x_n$ which is defined on a unit $n$-cube $I_n = [0, 1]^n$. Let $A_n$ be the integral of this function over $I_n$ (thus $A_n$ is the average value of this function on $I_n$). Now consider a function $g(x) := f(x, x, ..., x)$. It is defined on the unit interval $I_1$. Let $B$ be the integral of the function $g(x)$ over $I_1$.When it is true that $A_n < B$?

This general problem was motivated by the following specific question which is of great interest to me. Consider a function $f(x_1,x_2)= 1/(1+a_1x_1+a_2x_2)$ defined on a unit square $I_2=[0,1]^2$. It depends on two real parameters, $a_1$ and $a_2$ (ensuring that the denominator in the expression for the function $f$ is non-zero). Let $A_2$ be the integral of this function over $I_2$, and let $B$ be the integral of $g(x):= f(x,x)$ over the unit interval $I_1$. When the inequality $A_2 < B$ holds?

Having performed some tedious calculations I was able to investigate this problem in 2 special cases: $a_1=a_2$ (the inequality $A_2 < B$ holds) and $a_1=-a_2$ (the opposite inequality holds). I wonder if the analysis could be performed using some general ideas (using perhaps such properties as convexity upward/downward or monotonicity). Do you know if there is some general theory/results studying this type of problems?

Thank you very much for reading my question.

Answer 1

The approach suggested by Fedor is interesting but I see an issue with it. Namely:

The integral (over the square $ I_2 =\{(x, y): x, y \in [0, 1]\} $) of the left-hand side of the inequality listed by Fedor equals $2A_2$ (in my notation I follow Fedor, and instead of using $x_1$ and $x_2$ I use $x$ and $y$). But is the integral of the right-hand side over this square equal $2B$?

The integral $B$ is defined as the integral of $1/[1+ (a_1+a_2)x]$ from $x = 0$ to $x=1$. My calculation of this integral gives $B=(a_1+a_2)^{-1}\ln(1+a_1+a_2)$.

On the other hand, when we integrate the right-hand side of this inequality over $I_2$, we integrate $1/[1+(a_1+a_2)(x+y)/2]$. And this definite integral, according to my calculation, is not equal to $2B$.

So I think the original question is still open.

Answer 2

(New version, previous was completely wrong.)

Let us compare $f(x,y)+f(y,x)=g(a_1x+a_2y)+g(a_2x+a_1y)$ and $f(x,x)+f(y,y)=g(a_1x+a_2x)+g(a_1y+a_2y)$, where $g(t)=1/(1+t)$. Since $g$ is convex on $(0,+\infty)$ (I assume that $1+\min(a_1,a_2,a_1+a_2)>0$, else integrals diverge), comparing $g(A)+g(B)$ and $g(C)+g(D)$ with $A+B=C+D$ is equivalent to comparing $\max(A,B)$ and $\max(C,D)$. That is, if $a_1$ and $a_2$ have the same sign, we get $f(x,x)+f(y,y)\geqslant f(x,y)+f(y,x)$, otherwise $f(x,x)+f(y,y)\leqslant f(x,y)+f(y,x)$. Integrating over the square we compare your integrals.