Minimize $\displaystyle \prod_{i=1}^{2017}x_i$.

If $x_1,x_2,...,x_{2017} \in\Bbb{R}^+$ and $\displaystyle \dfrac{1}{1+x_1}+\dfrac{1}{1+x_2}+...+\dfrac{1}{1+x_{2017}} = 1$, find the minimal possible value of $\displaystyle \prod_{i=1}^{2017}x_i$.

Solution:

By cyclic symmetry, we know the critical value is at the point:

$\displaystyle \left(x_1,\cdots,x_{2017}\right)=(2016,\cdots,2016)$

And the objection function at that point is:

$\displaystyle f(2016,\cdots,2016)=2016^{2017}$

Now, looking at another point on the constraint:

$\displaystyle \left(4032,4032,\cdots,4032,\frac{2016}{2017}\right)$

We find the objective function at that point is:

$\displaystyle f\left(4032,4032,\cdots,4032,\frac{2016}{2017}\right)=\frac{2^{2016}2016^{2017}}{2017}>2016^{2017}$

And so we conclude:

$\displaystyle f_{\min}=2016^{2017}$