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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.

user164403
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    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 04 '14 at 20:52
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    Please consider accepting an answer by pressing the tick when you are happy. I realized you have not yet accepted any answer. – Lost1 Aug 04 '14 at 21:04
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    @ThomasAndrews Also, "question" is always redundant, which left the original title with the content of "an inequality". Hope the new one is better. –  Aug 04 '14 at 22:45

2 Answers2

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Hint: $$\frac{a_k}{(1+a_1)\cdots(1+a_k)} = \frac{1}{(1+a_1)\cdots(1+a_{k-1})} - \frac{1}{(1+a_1)\cdots(1+a_k)}$$

Thomas Andrews
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  • ... so we just have to prove that $$\prod_{i=1}^{n}(1+a_i)\geq 2^n,$$ right? – Jack D'Aurizio Aug 04 '14 at 20:58
  • Quite right. I left that step out... – Thomas Andrews Aug 04 '14 at 21:08
  • Could you let me know the name of that formula pls. – RealityDysfunction Aug 07 '14 at 02:26
  • It doesn't have a name. @RealityDysfunction – Thomas Andrews Aug 07 '14 at 02:32
  • 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 07 '14 at 02:51
  • @Thomas Andrews, how high-level is this question? I guess that is pretty vague, but I'd just like to know how deep into inequality proofs this is. I've started with learning some AM-GM and such. This was a question I encountered on the past Canadian Math Olympiad questions, and I just wanted to give a go as to how close I was to starting to solve Olympiad-level problems. Also, any suggestions on books that teach how to solve these type of problems? (Ones like these that I posted). Thanks. – user164403 Aug 07 '14 at 19:05
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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 $$1 - \frac{1}{b_n} \geq 1 - \frac{1}{2^n} = \frac{2^n-1}{2^n}.$$

André Nicolas
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  • Just wondering, what do you mean "by telescopy"? I only know of telescoping sums that cancel out, not so sure of what just telescopy is. Much thanks for the answer though. – user164403 Sep 06 '14 at 04:17
  • It is fairly standard jargon that covers telescoping sums and telescoping products. – André Nicolas Sep 06 '14 at 04:33