N.B: before downvoting this question without having finished reading it please leave a comment explaining why, so I can improve and ask better question in future, thanks.
In a problem I've to proof that $$\sum_{x_i\in A}{x_i}=1$$ After some effort I found the follow identity $$\sum_{x_i\in A}{x_i}=\sum_{k=max(0,n-b)}^{min(a,n)}{\frac{\binom{a}{k}\binom{b}{n-k}}{\binom{b+a}{n}}}$$ Whit $a\ge1, b\ge1, 1\le n\le a+b$
I'd like to bound this sum between itself calculated for $n=1$ and $n=a+b$ in this way:
$$\sum_{k=max(0,1-b)}^{min(a,1)}{\frac{\binom{a}{k}\binom{b}{1-k}}{\binom{b+a}{1}}}\le\sum_{k=max(0,n-b)}^{min(a,n)}{\frac{\binom{a}{k}\binom{b}{n-k}}{\binom{b+a}{n}}}\le\sum_{k=max(0,a+b-b)}^{min(a,a+b)}{\frac{\binom{a}{k}\binom{b}{a+b-k}}{\binom{b+a}{a+b}}}$$
If this inequality is true then
$$1\le\sum_{k=max(0,n-b)}^{min(a,n)}{\frac{\binom{a}{k}\binom{b}{n-k}}{\binom{b+a}{n}}}\le1$$ So $$\sum_{k=max(0,n-b)}^{min(a,n)}{\frac{\binom{a}{k}\binom{b}{n-k}}{\binom{b+a}{n}}}=1$$
Because
$$\sum_{k=max(0,1-b)}^{min(a,1)}{\frac{\binom{a}{k}\binom{b}{1-k}}{\binom{b+a}{1}}}=\sum_{k=0}^{1}{\frac{\binom{a}{k}\binom{b}{1-k}}{\binom{b+a}{1}}}={\frac{\binom{a}{0}\binom{b}{1}}{\binom{b+a}{1}}}+{\frac{\binom{a}{1}\binom{b}{0}}{\binom{b+a}{1}}}=\frac{a+b}{b+a}=1$$
and
$$\sum_{k=max(0,a+b-b)}^{min(a,a+b)}{\frac{\binom{a}{k}\binom{b}{a+b-k}}{\binom{b+a}{a+b}}}=\sum_{k=a}^{a}{\frac{\binom{a}{k}\binom{b}{a+b-k}}{\binom{b+a}{a+b}}}=\binom{a}{a}\binom{b}{a+b-a}=1$$
The problem is that i can't check this inequality my myself, can anyone explain me how to do it?
All you need is a double counting argument. If we select $n$ objects from a set of $a+b$ objects there are $\binom{a+b}{n}$ ways to do this. On the other hand, if we pick $n$ objects from $a+b$ objects by selecting some $k$ objects from $a$ and the remainder from $b$, then if we sum over all possible values of $k$, we count the number of ways of selecting $n$ objects from $a+b$ objects. Thus $$ \binom{a+b}{n}=\sum_{k=\max(0,n-b)}^{\min(a,n)} \binom{a}{k}\binom{b}{n-k}. $$
Now divide both sides by $\binom{a+b}{n}$ and use the identity you have already proved to get your result.