New bound for Am-Gm of 2 variables

Today I'm interested by the following problem :

Let $x,y>0$ then we have : $$x+y-\sqrt{xy}\leq\exp\Big(\frac{x\ln(x)+y\ln(y)}{x+y}\Big)$$

The equality case comes when $x=y$

My proof uses derivative because for $x\geq y $ the function : $$f(x)=x+y-\sqrt{xy}-\exp\Big(\frac{x\ln(x)+y\ln(y)}{x+y}\Big)$$

is decreasing and for $y\geq x$ the function is increasing and the maximum occurs when $x=y$

My question is : Have you an alternative proof wich doesn't use derivative ?

Thanks in advance.

This is not an answer to the question, but it's too big to put it to the comment. I will show some connection (which might be interesting) between this inequality and Shannon entropy.

Firstly rewrite this expression as $$ x + y- \sqrt{xy} \le x^{\frac{x}{x+y}} \cdot y^{\frac{y}{x+y}}. $$ Then one can use a substitution $$ a = \frac{x}{x+y}, \; b = \frac{y}{x+y} $$ and get an equivalelent inequality in terms of $a$ and $b$ $$ (1 - \sqrt{ab}) \le a^a b^b, \;\; a +b = 1. $$ So, we need to show that given $a+b = 1$, we will have $$ \sqrt{ab} + a^a b^b \ge 1. $$ It's equivalent to the following upper bound for Shannon entropy $H(a,b)$ $$ H(a,b) = -a \log a - b\log b \le -\log(1-\sqrt{ab}), \;\; a+b=1. $$

So, one needs to prove this estimate for Shannon entropy $H(a,b)$. Unfortunately, I have no idea how to do this without calculus. Plots of $H(a,b)$ and its upper bound:

It might happen that this inequality has some meaning in information theory, though I haven't found anything about that.

Incomplete answer

This is a trick that sometimes works when dealing with inequalities with two variables; however, in this case, the prohibition of calculus makes the problem more difficult.

Let $y=ax$ for some $a,x>0$. Then \begin{align}x+y-\sqrt{xy}\leq\exp\left(\frac{x\ln x+y\ln y}{x+y}\right)&\impliedby x+ax-x\sqrt a\le\exp\left(\frac{x\ln x+ax\ln ax}{x+ax}\right)\\&\impliedby x(1-\sqrt a+a)\le\exp\left(\ln x+\frac{a\ln a}{1+a}\right)\\&\impliedby1-\sqrt a+a\le a^{\frac a{1+a}}\end{align} so it suffices to show that $(1-\sqrt a+a)^{a+1}\le a^a$ for all $a\in(0,1)$, where $y<x$ without loss of generality.

It may be worth noting that the inequality is extremely tight which can be seen via this visualisation.

A COMMENT.

Actualy holds the following (somewhat) conjecture generalization

If $N>1$ and $x_1,x_2,\ldots,x_N>0$, then $$ \frac{1}{N}\sum^{N}_{k=1}x_k-\sqrt[N]{\prod^{N}_{k=1}x_k}\leq\exp\left(\frac{\sum^{N}_{k=1}x_k\log x_k}{\sum^{N}_{k=1}x_k}\right)\tag 1 $$