Inequality $f_{n}(x)>1$ for $0Let $0<x< 1$ and $n\geq 1$ a positive natural number then we have :
$$f_{n}(x)=\left(x\left(\frac{1-x}{1+x}\right)^n\right)^{\left(\frac{x}{x+1}\left(\frac{1-x}{1+x}\right)^n\right)}-x\left(\frac{1-x}{1+x}\right)^n+\sqrt{x\left(\frac{1-x}{1+x}\right)^n}>1$$
I  can show a related result wich is a limit because we have :
$$\lim_{n\to\infty}\left(x\left(\frac{1-x}{1+x}\right)^n\right)^{\left(\frac{x}{x+1}\left(\frac{1-x}{1+x}\right)^n\right)}-x\left(\frac{1-x}{1+x}\right)^n+\sqrt{x\left(\frac{1-x}{1+x}\right)^n}=1$$
But it doesn't help so much to solve the inequality .It just show that the inequality is sharper and sharper as $n\to \infty$.
The base case :
Using Wolfram alpha we have :
$$f_{1}(x)=1+\frac{\sqrt{1-x}}{\sqrt{2}}+\frac{1}{4}(x-1)(-\log(1-x)+2+\ln(2))+O((x-1)^{1.5})
$$
Unfortunately it's not sufficient to show the base case .
Edit :For the base case a better approximation is here
Also we can use  Newton's expansion (so without derivative) to kill the exponent for the general case.Remains to replace $a$ by $x$.It's around $1$ and we can also use an expansion around $0$.
Question :
How to show that $f_{n}(x)>1$ ?
I'm thanksful to anyone who have interest for the question and want to share something with me and other person on this subject.
 A: Here is $f_n$ drawn by numpy. Some numerical instability near $x=0$, but the variations of $f_n$ look fairly simple.
Proof
Let
\begin{equation}{u}_{n} \left(x\right) = x {\left(\frac{1-x}{1+x}\right)}^{n}\end{equation}
Obviously for $x \in  \left]0 , 1\right[$ we have
$ {u}_{n} \in  \left]0 , 1\right[$ and
for $n  \geqslant  1$ one has $ {u}_{n} \left(x\right)  \leqslant  {u}_{1} \left(x\right)$.
Using the inequality $ {e}^{y}  \geqslant  1+y$, we estimate
\begin{equation}\begin{array}{rcl}{f}_{n} \left(x\right)&=&\displaystyle  {e}^{{u}_{n} \log  \left({u}_{n}\right)/\left(x+1\right)}-{u}_{n}+\sqrt{{u}_{n}}\\
& \geqslant &\displaystyle  1+\frac{{u}_{n} \log  \left({u}_{n}\right)}{x+1}-{u}_{n}+\sqrt{{u}_{n}}
\end{array}\end{equation}
Let now $ {\gamma} \left(u\right) =-\sqrt{u} \log  \left(u\right)$ for
$ u \in  \left]0 , 1\right[$.
By differentiating $ {\gamma}$, we see that it admits a maximum for
$ u = {e}^{{-2}}$ and we deduce $ {\gamma} \left(u\right)  \leqslant  c = 2 {e}^{{-1}}$. It follows that $ u \log  \left(u\right)  \geqslant -c \sqrt{u}$. We use this to
further estimate $ {f}_{n}$:
\begin{equation}\renewcommand{\arraystretch}{2}  \begin{array}{rcl}{f}_{n} \left(x\right)& \geqslant &\displaystyle  1-c \frac{\sqrt{{u}_{n}}}{x+1}-{u}_{n}+\sqrt{{u}_{n}}\\
& \geqslant &\displaystyle  1+\sqrt{{u}_{n}} \left(\left(1-\frac{c}{1+x}\right)-\sqrt{{u}_{n}}\right)\\
& \geqslant &\displaystyle  1+\sqrt{{u}_{n}} \left(\left(1-\frac{c}{1+x}\right)-\sqrt{{u}_{1}}\right)
\end{array}\end{equation}
Thus, it suffices to show that
\begin{equation}\sqrt{{u}_{1}}  <  1-\frac{c}{1+x}\end{equation}
This reduces to the cubic inequation
\begin{equation}P \left(x\right) = {x}^{3}+{x}^{2}-\left(2 c-1\right) x+{\left(1-c\right)}^{2}  >  0\end{equation}
We see that $ {P'} \left(x\right) = 3 {x}^{2}+2 x-\left(2 c-1\right)$ and it implies that
$P$ has a unique minimum in $\left]0 , 1\right[$,
reached at $ {x}_{0} = \left(\sqrt{6 c-2}-1\right)/3$. We remark that
$6 c-2 = 12 {e}^{{-1}}-2  \geqslant  12/3-2 = 2$.
It only remains to prove
that $P \left({x}_{0}\right)  >  0$. We note that $ {P'} \left({x}_{0}\right) = 0$ implies
$\left(2 c-1\right) {x}_{0} = 3 {x}_{0}^{3}+2 {x}_{0}^{2}$, hence
\begin{equation}P \left({x}_{0}\right) = {\left(1-c\right)}^{2}-2 {x}_{0}^{3}-{x}_{0}^{2}\end{equation}
We note that $ e  >  2.67  >  8/3 \Longrightarrow  c = 2 {e}^{{-1}}  <  3/4$.
Hence $ 1-c  >  1/4$ and $ {\left(1-c\right)}^{2}  >  1/16 = 6/96  >  0.06$.
We also have $ {x}_{0}  <  0.2$ because $ {P'} \left(0.2\right) = 1.52-2 c  >  2 \left(3/4-c\right)  >  0$. Hence $ 2 {x}_{0}^{3}+{x}_{0}^{2}  <  0.056  <  0.06$. Hence $P \left({x}_{0}\right)  >  0$, QED.
Let us conclude with a graphical comparison between $\sqrt{{u}_{n}}$ and
$\displaystyle  1-\frac{c}{1+x}$:

A: Just partial results.
$\forall n >1$, $f_n(0)=f_n(1)=1$.
Close to $x=0$, there is no problem since
$$f_n(x)=1+x^{1/2}+x (\log (x)-1)-n x^{3/2}+O\left(x^{2}\right)$$
For $x=\frac 12$, we have
$$f_n\left(\frac{1}{2}\right)=-\frac{3^{-n}}{2}+\frac{3^{-n/2}}{\sqrt{2}}+2^{-3^{-n-1}} 3^{-3^{-n-1} n}\quad > ~~1 \quad \forall n$$
Close to $x=1$, I have been unable to build the series expansion (because of the first term).
Edit
Let $x=\tan^2(y)$ to make
$$g_n(y)=\tan (y) \left(\sqrt{\cos ^n(2 y)}-\tan (y) \cos ^n(2 y)\right)+\left(\tan ^2(y)
   \cos ^n(2 y)\right)^{\sin ^2(y) \cos ^n(2 y)}$$ So, around $y=0$
$$g_n(y)=1+y+y^2 (2 \log (y)-1)+\left(\frac{1}{3}-n\right) y^3+O\left(y^4\right)$$
$$g_n\left(\frac{\pi}{8}\right) \quad > ~~1 \quad \forall n$$
A: Observe the constant repeat of $ \frac {1-x}{1+x} $ in the equation.
$$0 < x < 1 \implies 0< \frac {1-x}{1+x} <1 \implies 0< (\frac {1-x}{1+x})^n <1$$
as $\frac {1-x}{1+x}$ is strictly decreasing. Let $t=(\frac {1-x}{1+x})^n$,
$$f(x)=(xt)^\frac{xt}{1+x}-xt+\sqrt {xt} $$
Note $0<xt<1$. Thus $0<\frac {xt}{1+x}<1$ as well.
To minimise the function, $\frac {xt}{1+x}$ should be as close to $1$ as possible, so as to minimise the value of $(xt)^\frac{xt}{1+x}$. Taking $\frac {xt}{1+x}=1$,
$$f(x)=xt-xt+\sqrt{xt}=\sqrt(x+1)>1$$
Not very rigorous, and I'm not even sure if it's very accurate, but here's my attempt. Do hope it helps.
