Pretty conjecture $x^{left(frac{y}{x}right)^n}+y^{left(frac{x}{y}right)^n}leq 1$

Question

inspired (again) by an inequality of Vasile Cirtoaje I propose my own conjecture :

Let $x,y>0$ such that $x+y=1$ and $ngeq 1$ 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 $a^{(frac{a}{b})^p}+b^{(frac{b}{a})^p}+cgeq 1$
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 ($kgeq 2$):
$$x^{left(frac{1-x}{x}right)^{2k}}+(1-x)^{left(frac{x}{1-x}right)^{2k}}leq (p(x)+varepsilon)^{left(frac{1-x}{x}right)^{2k-2}}+q(x)^{left(frac{x}{1-x}right)^{2k-2}}$$
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 theoreticalpractical method ?

River Li · Answer

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 $z ge frac{33}{14}$,
$$zln(z) - ln(z - 1) - 2ln z ge 0$$
which is true (easy).

EdG · Answer

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 $qgeq 1$ (you can do everything in reverse with $x geq 0.5$ and $qleq 1$). 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 $b_ngeq0.5$, 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
$$
(1-b_n)^q+b_n^{1/q} leq 1, quad forall ,,, b_n in [0.5, 1], qgeq1 
$$
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 $1$.

Pretty conjecture $x^{left(frac{y}{x}right)^n}+y^{left(frac{x}{y}right)^n}leq 1$

Little update

Theoretical method

My second question

2 Answers

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