How prove this $\prod_{i=1}^{n}a_{i}+\prod_{i=1}^{n}(1-a_{i})\ge\frac{1}{2^{n-1}}$ let $0<\le a_{i}\le \dfrac{1}{2},i=1,2,\cdots,n$.show that
$$\prod_{i=1}^{n}a_{i}+\prod_{i=1}^{n}(1-a_{i})\ge\dfrac{1}{2^{n-1}}$$
my idea: I guess this problem will use Bernoulli inequality：
$$(1+x_{1})(1+x_{2})\cdots (1+x_{n})\ge 1+x_{1}+x_{2}+\cdots+x_{n}$$
where $x_{i}\ge -1$
But I can't show it, and it say that can  induction? maybe this problem have other nice methods,Thank you everyone.

Let $a_i=\frac{1}{2}-x_i.$ Then our inequality can be rewritten as
$$\prod(\frac{1}{2}-x_i)+\prod(\frac{1}{2}+x_i)\ge \frac{1}{2^{n-1}}.$$
Now simple induction on $n$ works.
@math110: Finishing induction: so the base case is clear. Now assuming that the inequality is true for $n$ variables we would like to prove
$$\prod_1^{i=n+1}(\frac{1}{2}-x_i)+\prod_{i=1}^{n+1}(\frac{1}{2}+x_i)\ge \frac{1}{2^{n}}.$$ Indeed,
$$\prod_{i=1}^{n+1}(\frac{1}{2}-x_i)+\prod_{i=1}^{n+1}(\frac{1}{2}+x_i)$$ $$=\frac{1}{2}\left(\prod_{i=1}^{n}(\frac{1}{2}-x_i)+\prod_{i=1}^{n}(\frac{1}{2}+x_i)\right)$$ $$+x_{n+1}\left(\prod_{i=1}^{n}(\frac{1}{2}+x_i)-\prod_{i=1}^{n}(\frac{1}{2}-x_i)\right)\ge \frac{1}{2}\cdot\frac{1}{2^{n-1}}+0=
\frac{1}{2^n}$$
