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Let $x_1x_2…x_n$ be positive real numbers such that $ $$\tfrac{1}{1+x_1}$$ + $$\tfrac{1}{1+x_2}$$ +… + \tfrac{1}{1+x_n}=1 $

Prove that $x_1x_2…x_n \ge (n-1)^n$

Please can someone help with this question using a proven inequality such as AM GM or Cauchy Schwarz i recieved the problem at a maths olympiad camp for practice ive solved inequalities before but Im struggling to solve this one.

Tyrone
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    I think you forgot the dollar signs around the math to turn it into mathjax – TrostAft Dec 15 '18 at 17:38
  • It seems GM-HM could be useful since you can have the following$$\sqrt[n]{(x_1+1)(x_2+1)...(x_n+1)} \ge \frac{n}{\frac{1}{(x_1+1)}+\frac{1}{(x_2+1)}+...+\frac{1}{(x_n+1)}}=n$$ i.e. $$(x_1+1)(x_2+1)...(x_n+1) \ge n^n$$ which isn't what we're looking for but it's fairly similar – Spasoje Durovic Dec 15 '18 at 18:10
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    Hello and welcome to math.stackexchange. Please write about the context: Where does this problem come from? What tools do you know? Can you solve similar problems, e.g. if $n=2$ or if the fractions are $\frac{1}{x_i}$? – Hans Engler Dec 15 '18 at 18:10
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    The thing you want to prove is equivalent to $1+\sqrt[n]{x_1x_2\cdots x_n}\ge \cfrac n{\cfrac 1{1+x_1}+\cfrac 1{1+x_2}+\cdots+\cfrac 1{1+x_n}}$. This is finer than AM-GM. Jensen Inequality might work. Refer to this https://math.stackexchange.com/questions/1600051/prove-that-frac11x-1-frac11x-2-cdots-frac11x-n-geq-fracn?rq=1 – Lance Dec 15 '18 at 18:21
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    Possible duplicate of Prove the following inequality $\sqrt[n]{x_1x_2...x_n}\ge (n-1)$ – Surb Dec 15 '18 at 18:35
  • @Lance Unfortunately, this inequality doesn't help here. In your link, it is assumed that $x_1,x_2,\ldots,x_n\ge 1$. This is not an assumption in this thread's problem. –  Dec 15 '18 at 18:36
  • @Snookie That's why I used a 'might' :). Surb's link is very interesting. It seems solvable by creatively using AM-GM. – Lance Dec 15 '18 at 18:39
  • You want to prove that the product is minimized when all the $x_i$ are equal. This is easy with Lagrange multipliers, though it doesn't use "a proven inequality." – saulspatz Dec 15 '18 at 18:58

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