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.

  • $\begingroup$ Hello,$a_{i}=0.3?$ $\endgroup$ – math110 Sep 5 '13 at 3:00
  • $\begingroup$ if all $a_i$ are the same then the inequality follows from Jensen's for $f(x)=x^n.$ $\endgroup$ – leshik Sep 5 '13 at 3:02
  • $\begingroup$ yes,@leshik,But this not same $\endgroup$ – math110 Sep 5 '13 at 3:04

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

$$ 2^{n} \left( \prod a_i + \prod (1-a_i) \right) \geq 2 \prod ( 1 - a_i + a_i) = 2 .$$

  • $\begingroup$ It's nice,Thank you +1 $\endgroup$ – math110 Sep 5 '13 at 7:23

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, $$\prod_{i=1}^{n+1}(\frac{1}{2}-x_i)+\prod_{i=1}^{n+1}(\frac{1}{2}+x_i)$$ $$=\frac{1}{2}\left(\prod_{i=1}^{n}(\frac{1}{2}-x_i)+\prod_{i=1}^{n}(\frac{1}{2}+x_i)\right)$$ $$+x_{n+1}\left(\prod_{i=1}^{n}(\frac{1}{2}+x_i)-\prod_{i=1}^{n}(\frac{1}{2}-x_i)\right)\ge \frac{1}{2}\cdot\frac{1}{2^{n-1}}+0= \frac{1}{2^n}$$

  • $\begingroup$ Thank you,I consider it.can you post your following induction? $\endgroup$ – math110 Sep 5 '13 at 3:01

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.

  • $\begingroup$ And the base case holds. $\endgroup$ – Did Sep 5 '13 at 15:21

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