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.

As @leshik suggests, one can proceed by induction.
Suppose this is true for $n$, and add an additional term $a_{n+1}$.
Let $P = \prod\limits_{i=1}^{n} a_i$ and $Q = \prod\limits_{i = 1}^n (1-a_i)$.
Then
\begin{align*}
\prod_{i=1}^{n + 1} a_i + \prod_{i = 1}^{n+1} (1 - a_i)
&= a_{n+1} P + (1-a_{n+1}) Q \\
&= \frac12 \left(P + Q\right) + \left(\frac12 - a_{n + 1}\right)(Q - P)
\end{align*}
By induction hypothesis $P + Q \ge \frac{1}{2^{n-1}}$, and clearly $Q \ge P$ since $a_i \le 1 - a_i$ for all $i$.  Thus the above is greater than or equal to
$$
\frac12 \left( \frac{1}{2^{n-1}} \right) + 0 = \frac{1}{2^n}
$$
as desired.
