MHB Find Min Value of Expression: $\prod_{i=1}^{2017}x_i$

[lfdahl]

If $x_1,x_2,...,x_{2017} \in\Bbb{R}_+$
and $\frac{1}{1+x_1}+\frac{1}{1+x_2}+...+\frac{1}{1+x_{2017}} = 1$
- then find the minimal possible value of the expression: \[\prod_{i=1}^{2017}x_i\]

 

[MarkFL]

My solution:

Spoiler

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

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

And the objection function at that point is:

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

Now, looking at another point on the constraint:

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

We find the objective function at that point is:

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

And so we conclude:

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

 

[lfdahl]

My solution:

Spoiler

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

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

And the objection function at that point is:

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

Now, looking at another point on the constraint:

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

We find the objective function at that point is:

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

And so we conclude:

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

Thankyou, MarkFL for your correct solution!