# If $a_1a_2\cdots a_n=1$, then the sum $\sum_k a_k\prod_{j\le k} (1+a_j)^{-1}$ is bounded below by $1-2^{-n}$

I am having trouble with an inequality. Let $a_1,a_2,\ldots, a_n$ be positive real numbers whose product is $1$. Show that the sum $$\frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\frac{a_3}{(1+a_1)(1+a_2)(1+a_3)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)}$$

is greater than or equal to $$\frac{2^n-1}{2^n}$$

If someone could help approaching this, that would be great. I don't even know where to start.

• Generally, words like "tough," "hard," "difficult" aren't very useful in the title, because they give no real information. Presumably when you ask the question here, it is because you find it hard. But does it let a potential answerer know if they can help you? – Thomas Andrews Aug 4 '14 at 20:52
• Please consider accepting an answer by pressing the tick when you are happy. I realized you have not yet accepted any answer. – Lost1 Aug 4 '14 at 21:04
• @ThomasAndrews Also, "question" is always redundant, which left the original title with the content of "an inequality". Hope the new one is better. – user147263 Aug 4 '14 at 22:45

Note that for that every positive integer $i$ we have \begin{eqnarray} \frac{a_i}{(1+a_1)(1+a_2) \cdots (1+a_i)} & = & \frac{1 + a_i}{(1+a_1)(1+a_2) \cdots (1+a_i)} - \frac{1}{(1+a_1)(1+a_2) \cdots (1+a_i)} \nonumber \\ & = & \frac{1}{(1+a_1) \cdots (1+a_{i-1})} - \frac{1}{(1+a_1) \cdots (1+a_i)}. \nonumber \end{eqnarray} Let $b_i = (1+a_1)(1+a_2) \cdots (1+a_i)$, with $b_0= 0$. Then by telescopy $$\sum\limits_{i=1}^n \left( \frac{1}{b_{i-1}} - \frac{1}{b_i} \right) = 1 - \frac{1}{b_n}.$$ Since $1+x\geq 2\sqrt{x}$ for all $x\ge 0$, we have $$b_n = (1+a_1)(1+a_2) \cdots (1+a_n) \geq (2 \sqrt{a_1})(2 \sqrt{a_2}) \cdots (2 \sqrt{a_n}) = 2^n,$$ with equality precisely if $a_i=1$ for all $i$. It follows that $$1 - \frac{1}{b_n} \geq 1 - \frac{1}{2^n} = \frac{2^n-1}{2^n}.$$
Hint: $$\frac{a_k}{(1+a_1)\cdots(1+a_k)} = \frac{1}{(1+a_1)\cdots(1+a_{k-1})} - \frac{1}{(1+a_1)\cdots(1+a_k)}$$
• ... so we just have to prove that $$\prod_{i=1}^{n}(1+a_i)\geq 2^n,$$ right? – Jack D'Aurizio Aug 4 '14 at 20:58
• It's just $$1-\frac{1}{a_k}=\frac{a_k}{1+a_k}$$ multiplied on both sides by $$\frac{1}{(1+a_1)(1+a_2)\dots(1+a_{k-1})}$$ – Thomas Andrews Aug 7 '14 at 2:51