Pretty conjecture $x^{\left(\frac{y}{x}\right)^n}+y^{\left(\frac{x}{y}\right)^n}\leq 1$ inspired (again) by an inequality of Vasile Cirtoaje I propose my own conjecture :

Let $x,y>0$ such that $x+y=1$ and $n\geq 1$ a natural number then we have :
$$x^{\left(\frac{y}{x}\right)^n}+y^{\left(\frac{x}{y}\right)^n}\leq 1$$

First I find it very nice because all the coefficient are $1$ .
I have tested with Geogebra until $n=50$ without any counter-examples.
Furthermore we have an equality case as $x=y=0.5$ or $x=1$ and $y=0$ and vice versa .
To solve it I have tried all the ideas here
My main idea was to make a link with this inequality (my inspiration) see here
So if you can help me to solve it or give me an approach...
...Thanks for all your contributions !
Little update
I think there is also an invariance as in question here Conjecture $a^{(\frac{a}{b})^p}+b^{(\frac{b}{a})^p}+c\geq 1$
Theoretical method
Well,Well this method is very simple but the result is a little bit crazy (for me (and you ?))
Well ,I know that if we put $n=2$ we can find (using parabola) an upper bound like
$$x^{\left(\frac{1-x}{x}\right)^2}\leq ax^2+bx+c=p(x)$$
And
$$(1-x)^{\left(\frac{x}{1-x}\right)^2}\leq ux^2+vx+w=q(x)$$
on $[\alpha,\frac{1}{2}]$ with $\alpha>0$ and such that $p(x)+q(x)<1$
In the neightborhood of $0$ we can use a cubic .
Well,now we have (summing) :
$$x^{\left(\frac{1-x}{x}\right)^2}+(1-x)^{\left(\frac{x}{1-x}\right)^2}\leq p(x)+q(x)$$
We add a variable $\varepsilon$ such that $(p(x)+\varepsilon)+q(x)=1$
Now we want an inequality of the kind ($k\geq 2$):
$$x^{\left(\frac{1-x}{x}\right)^{2k}}+(1-x)^{\left(\frac{x}{1-x}\right)^{2k}}\leq (p(x)+\varepsilon)^{\left(\frac{1-x}{x}\right)^{2k-2}}+q(x)^{\left(\frac{x}{1-x}\right)^{2k-2}}$$
Now and it's a crucial idea we want something like :
$$\left(\frac{x}{1-x}\right)^{2k-2}\geq \left(\frac{1-(p(x)+\varepsilon)}{q(x)}\right)^y$$
AND :
$$\left(\frac{1-x}{x}\right)^{2k-2}\geq \left(\frac{1-q(x)}{p(x)+\varepsilon}\right)^y$$
Now it's not hard to find a such $y$ using logarithm .
We get someting like :
$$x^{\left(\frac{1-x}{x}\right)^{2k}}+(1-x)^{\left(\frac{x}{1-x}\right)^{2k}}\leq q(x)^{\left(\frac{1-q(x)}{q(x)}\right)^{y}}+(1-q(x))^{\left(\frac{q(x)}{1-q(x)}\right)^{y}}$$
Furthermore the successive iterations of this method conducts to $1$ because the values of the differents polynomials (wich are an approximation of the initial curve) tend to zero or one (as abscissa).
The extra-thing (and a little bit crazy) we can make an order on all the values.
My second question
Is it unusable as theoretical\practical method ?
 A: I am sorry for not proving your conjecture, but I thought of writing up my thoughts as it might help you.
Without loss of generality, we can say that $y \geq 0.5$. Let $q=y/x$ and because $y \geq 0.5$ we have $q\geq 1$ (you can do everything in reverse with $x \geq 0.5$ and $q\leq 1$). Let $a_0=x$ and $b_0=y$. Then we can write:
$$
x^{\left( \frac{y}{x} \right)^n} + y^{\left(\frac{x}{y}\right)^n}=x^{q^n}+y^{\left(\frac{1}{q}\right)^n}=a_n+b_n,
$$
with
$$\begin{align}
a_n &= a_{n-1}^q,\\
b_n &= b_{n-1}^{1/q}.
\end{align}$$
It is easy to verify that with $b_n\geq0.5$, also $b_{n+1}\geq 0.5$. Now, let us assume that $a_{n-1} \leq 1-b_{n-1}$. For $n=1$, we have $a_0 \leq 1-b_0$ (more specifically, $a_0=1-b_0$). Because of that, we can write (remember that $q \geq 1$):
\begin{align}
a_{n+1} + b_{n+1} &= a_n^q+b_n^{1/q} \\
&\leq (1-b_n)^q + b_n^{1/q}.
\end{align}
What is left, is to prove that
$$
(1-b_n)^q+b_n^{1/q} \leq 1, \quad \forall \,\,\, b_n \in [0.5, 1], q\geq1 
$$
This is where I got stuck, but perhaps you know how to continue from here. In the figure below, I plotted $(1-b)^q+b^{1/q}$ with $b$ on the interval $[0.5,1]$ for various values of $q$ and it seems that the result is always equal or below $1$.

A: Partial answer
Let us prove the case when $0 < x \le \frac{14}{33}$.
By Bernoulli inequality, we have
$$y^{(x/y)^n} = (1 - x)^{(x/y)^n} \le 1 - x(x/y)^n.$$
It suffices to prove that
$$x^{(y/x)^n} + 1 - x(x/y)^n \le 1$$
or
$$\ln x \le \frac{- \ln[(y/x)^n]}{(y/x)^n - 1}.$$
Since $u \mapsto \frac{-\ln u}{u - 1}$
is strictly increasing on $u > 1$, it suffices to prove that
$$\ln x \le \frac{- \ln[(y/x)]}{(y/x) - 1}.$$
With the substitution $x = \frac{1}{z}$, it suffices to prove that, for all $z \ge \frac{33}{14}$,
$$z\ln(z) - \ln(z - 1) - 2\ln z \ge 0$$
which is true (easy).
