How to prove this double integral identity with simaler Jenson decomposition? Question:

let $x=(x_{1},x_{2},\cdots,x_{n})$,and $f:I\to R$have two derivative,and for any $[a,b]\subseteq I$,then $f''$ is integrable on $[a,b]$,and $x\in I^n,n\ge 2$
show that
$$J[f(x)]=\dfrac{1}{n^2}\sum_{1\le i<j\le n}\left(\int\int_{\Omega}f''[t_{1}x_{i}+t_{2}x_{j}+(1-t_{1}-t_{2})A(x)]dt_{1}dt_{2}\right)\cdot (x_{i}-x_{j})^2$$
where $$J[f(x)]=\dfrac{1}{n}\sum_{i=1}^{n}f(x_{i})-f\left(\dfrac{1}{n}\sum_{i=1}^{n}x_{i}\right),A(x)=\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}$$
$$\Omega=\{(t_{1},t_{2})|t_{1}\ge 0,t_{2}\ge 0,t_{1}+t_{2}\le 1\}$$

This problem is from Graduate courses when I read book,this  author say have simple  calculate,then we have this identity,But I try sometimes,and follow more and more ugly?can you help? and I fell this relsut is interesting,are you agree with me?
Thank you
 A: Lemma. Suppose that $f(0)=0$ and $x,y$ are nonzero numbers then
$$
(x-y)^2\int\!\!\int_\Omega f''(tx+sy)dtds=f(x)+f(y)-x\frac{f(y)}{y}-y\frac{f(x)}{x}
$$
Proof. Indeed, if
$$I=(x-y)^2\int\!\!\int_\Omega f''(tx+sy)dtds$$ then
$$\eqalign{
I&=(x-y)^2\int_0^1\left(\int_0^{1-s}f''(tx+sy)dt\right)ds\\
&=(x-y)^2\int_0^1\left[\frac{1}{x}f'(tx+sy)\right]_{t=0}^{t=1-s} ds\\
&=\frac{(x-y)^2}{x}\int_0^1\left( f'(x+s(y-x))-f'(sy)\right) ds\\
&=\frac{(x-y)^2}{x}\left[\frac{1}{y-x}f(x+s(y-x))-\frac{1}{y}f(sy)\right]_{s=0}^{s=1}\\
&=\frac{(x-y)^2}{x}\left(\frac{f(y)-f(x)}{y-x}-\frac{f(y)}{y}\right)\\
&=f(x)+f(y)-x\frac{f(y)}{y}-x\frac{f(y)}{y}
}$$
and the Lemma follows.
Remark. By continuity, if we interpret $\dfrac{f(t)}{t}$ as $f'(0)$ when $t=0$, we see that the lemma remains valid is $x$ or $y$ is zero. 
A Particular case. Assume first that $A(x)=0$, $f(0)=0$. In this case we have
$$
\eqalign{\sum_{1\leq i,j\leq n}(x_i-x_j)^2\int\!\!\int_\Omega f''(t x_i+sx_j)dtds
&=\sum_{1\leq i,j\leq n}(x_i-x_j)^2\left(f(x_i)+f(x_j)-x_i\frac{f(x_j)}{x_j}-x_j\frac{f(x_i)}{x_i}\right)\cr
&=2n\sum_{k=1}^nf(x_k)-2\left(\sum_{k=1}^n x_k\right)\left(\sum_{l=1}^n \frac{f(x_l)}{x_l}\right)\cr
&=2n\sum_{k=1}^nf(x_k)
}
$$
Because we assumed that $A(x)=0$. This proves the result when $f(0)=0$ and $A(x)=0$.
The general case. Given $(x_1,\ldots,x_n)\in I^n$, We just apply the particular case to the function
$$u\mapsto F(u)=f(u+A(x))-f(A(x))$$
and to $(X_1,\ldots,X_n)$ with $X_k=x_k-A(x)$, $k=1,\ldots,n$. The desired conclusion follows.
