Assume $0 < a_i\leq 1$ for $i = 1,2,...,n,$ and $$\sum_{i=1}^{n} {a_i} = a.$$ Show that $$\sum_{i=1}^{n} \frac {a_i}{1+a-a_i} + \prod_{i=1}^{n} {(1-a_i)} \leq 1.$$
I first substituted $b_i = 1-a_i, 0 < b_i\leq 1$ to simply the problem to $$\sum_{i=1}^{n} \frac {1-b_i}{a+b_i} + \prod_{i=1}^{n} {b_i} \leq 1.$$ Rearranging: $$\sum_{i=1}^{n} \frac {1-b_i}{a+b_i} \leq 1 - \prod_{i=1}^{n} {b_i}.$$ From here I thought about trying to telescope the sum and but didn't really get far, and I'm not sure how to deal with the product and sum in the same inequality.
Let $[n]:=\{1,2,\ldots,n\}$. Consider $$f\left(a_1,a_2,\ldots,a_n\right):=\sum_{i=1}^n\,\frac{a_i}{1+\sum_{j\in[n]\setminus\{i\}}\,a_j}+\prod_{i=1}^n\,\left(1-a_i\right)$$ for $a_1,a_2,\ldots,a_n\in[0,1]$. Observe that $f$ is convex in each variable. Therefore, the maximum value of $f$ is attained at the boundary. The task is now to show that $$f\left(a_1,a_2,\ldots,a_n\right)\leq 1$$ for all $a_1,a_2,\ldots,a_n\in\{0,1\}$.
Fix $\left(a_1,a_2,\ldots,a_n\right)\in\{0,1\}^n$. Let $Z:=\left\{i\in[n]\,|\,a_i=1\right\}$. Ergo, if $Z$ is nonempty, then we have $$f\left(a_1,a_2,\ldots,a_n\right)=\sum_{i\in Z}\,\frac{1}{|Z|}=1\,.$$ If $Z=\emptyset$, then $a_i=0$ for all $i=1,2,\ldots,n$, whence $$f\left(a_1,a_2,\ldots,a_n\right)=\prod_{i=1}^n\,1=1\,.$$ In fact, $f\left(a_1,a_2,\ldots,a_n\right)=1$ if and only if $a_1,a_2,\ldots,a_n\in\{0,1\}$.