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.
 A: 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,
$$\prod_{i=1}^{n+1}(\frac{1}{2}-x_i)+\prod_{i=1}^{n+1}(\frac{1}{2}+x_i)$$ $$=\frac{1}{2}\left(\prod_{i=1}^{n}(\frac{1}{2}-x_i)+\prod_{i=1}^{n}(\frac{1}{2}+x_i)\right)$$ $$+x_{n+1}\left(\prod_{i=1}^{n}(\frac{1}{2}+x_i)-\prod_{i=1}^{n}(\frac{1}{2}-x_i)\right)\ge \frac{1}{2}\cdot\frac{1}{2^{n-1}}+0=
\frac{1}{2^n}$$
A: 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.
A: Apply the rearrangement inequality $2^{n-1}$ times.
Specifically, let $B$ denote the $2^{n}$ binary strings of $n$ variables, for $b\in B$, let $b_i$ denote the position in the $i$th coordinate, then
$$\prod a_i + \prod (1-a_i) \geq \prod_{b_i = 1} a_i \prod_{b_i = 0} (1-a_i) + \prod_{b_i = 1} (1-a_i) \prod_{b_i=0} a_i.$$
Add the $2^{n}$ inequalities, we get that
$$ 2^{n} \left( \prod a_i + \prod (1-a_i) \right) \geq 2 \prod ( 1 - a_i + a_i) = 2 .$$
