Exponential function inequality with n terms

Question

Let $a>0, a\neq1$ and $x_1,x_2,..,x_n \in \mathbb{R}$ such that $$\sum_{k=1}^n a^{x_k}\leq a$$ and $a^{x_k}+a^{x_j}<1$, $\forall $ $k\neq j$

Prove that $$\prod_{k=1}^n(1-a^{x_k})\geqslant1-a$$

Any ideas?

Answer 1

First observe that $a^{x_k}>0$ so that inequality $a^{x_k}+a^{x_j}<1$ means $a^{x_k}<1$. After this observation the question is effectively reduced upon introducing the notation $a_i=a^{x_i}$ to the following one:

Given $0<a_i<1$ and $\sum a_i\le a$, prove $\prod(1-a_i)\ge 1-a$. As $\sum a_i\le a$ it suffices to prove: $$ \prod_{i=1}^n(1-a_i)\ge 1-\sum_{i=1}^na_i\tag{1}. $$

For $n=1$ the inequality is trivial. Let us demonstrate that if it is valid for $n-1$ it is valid for $n$ as well: $$ \prod_{i=1}^n(1-a_i)=(1-a_n)\prod_{i=1}^{n-1}(1-a_i)\stackrel{I.H.}{\ge} (1-a_n)\left(1-\sum_{i=1}^{n-1}a_i\right)\\=1-a_n-\sum_{i=1}^{n-1}a_i+ a_n\sum_{i=1}^{n-1}a_i\ge 1-\sum_{i=1}^{n}a_i. $$

As can be easily observed the equality is possible only if $n=1$, and $a_1=a$ (meaning $x_1=1$). In all other cases strict inequality is valid.

Answer 2

If $0 \leq t_j \leq 1$ then $1-\prod (1-t_j) \leq \sum t_j$. This can be proved by an easy induction argument. Since $0<a^{x_k} <1$ we are done!