If $\prod x_k=1$, then $\prod\frac{1+x_k}{2}\leq ( \frac{x_1+\cdots +x_n}{n})^{n-1}$ Despite many attempts, no one at StackOverflow has succeeded in solving that old question about proving a deceptively simple-looking inequality.
I propose now a weaker and slightly simpler inequality (no more squares) which may perhaps be easier to prove.
Let $x_1,x_2, \ldots ,x_n$ be positive numbers whose product is $1$. Prove or find a counterexample :
$$
\prod_{k=1}^n \big(\frac{1+x_k}{2}\big) \leq \bigg( \frac{x_1+x_2+x_3+ \ldots x_n}{n}\bigg)^{n-1}
$$
(this inequality follows from the old one by putting $x_i=a_i^2$ and using Cauchy-Schwarz).
 A: I am going to propose something that might work for the reduced 2-variable problem (see the comments above). We need to prove that given $xy^n=1$ then:$$f(x,y)=\left(\frac{x+n y}{n+1}\right)^n-\frac{1+x}{2}\left(\frac{1+y}{2}\right)^n\geq 0$$ Effectively we need to show $f(y^{-n},y)\geq 0$. Now consider the polynomial $y^{n^2}f(y^{-n},y)$. It has a double root at $y=1$.
Moreover it looks like its coefficients can be computed effectively. I suggest it can be shown that there are only 4 sign alternations for this polynomial. Hence using the Descartes' rule of signs the number of positive roots is either 2 (in which case we are done because of the double root at 1) or 4. 
If there are 4 roots the remaining 2 roots are both either bigger then 1 or smaller then 1. Hence we are done for at least one of the cases $y>1$ or $y<1$.
This may be enough to establish the general case.
A: As mentioned in comment, one can use equal variable method for $f(x) = \log \left(\frac{1+x}{2}\right)$. Note that 
$$g(x) = f'(1/x) = \frac{x}{x+1}$$
is strictly concave for $x > 0$. By corollary 1.6, the maximum of LHS is reached when $0 < x_1 \leq 1 \leq x_2 = \cdots = x_n = t$. Substitute $x_1 = \frac{1}{t^{n-1}}$, we have to prove
$$\left(\frac{(n-1)t + \frac{1}{t^{n-1}}}{n}\right)^{n-1} \ge \left(\frac{1+t}{2}\right)^{n-1}\left(\frac{1+\frac{1}{t^{n-1}}}{2}\right)$$
for $t \ge 1$. By AMGM applied to $(n-2)$ $(1+t)/2$, and $(1+t)(1+t^{-(n-1)})/4$, we see that 
$$RHS \leq \left(\frac{(n-2)\left(\frac{1+t}{2}\right) + \frac{1+t+t^{-(n-1)} + t^{-(n-2)}}{4}}{n-1}\right)^{n-1}$$
So it suffices to show that 
$$\frac{(n-1)t + \frac{1}{t^{n-1}}}{n} \ge \frac{(n-2)\left(\frac{1+t}{2}\right) + \frac{1+t+t^{-(n-1)} + t^{-(n-2)}}{4}}{n-1}$$
Clearing denominator and simplify,
$$\Leftrightarrow 4(n-1)^2 t + 4(n-1)t^{-(n-1)} \ge 2n(n-2)t + 2n(n-2) + n (1+t+t^{-(n-1)} + t^{-(n-2)})$$
$$\Leftrightarrow (2n^2-5n+4)t + (3n-4)t^{-(n-1)} \ge 2n^2 - 3n + nt^{-(n-2)} \hspace{5mm} (*)$$
We prove the last inequality now. AM-GM gives
$$\frac{n(n-2)}{(n-1)} t^{-(n-1)} + \frac{n}{n-1} \ge nt^{-(n-2)}$$
and
$$(2n^2-5n+4)t + \frac{2n^2-5n+4}{n-1}t^{-(n-1)} \ge (2n^2-5n+4)\frac{n}{n-1}$$
(*) follows from adding up the last two inequalities.
