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Let $x_1,x_2,\ldots,x_n > 0$ such that $\dfrac{1}{1+x_1}+\dfrac{1}{1+x_2}+\cdots+\dfrac{1}{1+x_n} = 1$. Prove that $x_1x_2\cdots x_n \geq (n-1)^n$.

It seems like we should be able to use AM-GM here since the expression we want to prove has $x_1x_2\cdots x_n$ in it. But I don't know how to deal with this $\dfrac{1}{1+x_1}+\dfrac{1}{1+x_2}+\cdots+\dfrac{1}{1+x_n} = 1$ expression.

Edit: the solution in the link you provided was not very clear. I am wondering if someone can provide a more clear solution.

Jacob Willis
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  • Yes it does seem to be solved in this other question. Is there are general principle that says that the extreem values in such problems occurs when $x_1=x_2=x_3=\cdots $? – Rene Schipperus Dec 29 '15 at 16:14
  • ^I don't think there is such a general principle. For instance, the extreme values of $x^3+y^3$ subject to the constraint $x^2+y^2 = 1$ occur at $(x,y) = (\pm 1,0)$ and $(0,\pm 1)$. – JimmyK4542 Dec 29 '15 at 21:17
  • I posted a new solution here (using the Lagrange Multipliers Method). – Pedro Dec 29 '15 at 21:19

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