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.
 A: 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.
A: 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: A COMMENT.
Actualy holds the following (somewhat) conjecture generalization
If $N>1$ and $x_1,x_2,\ldots,x_N>0$, then
$$
\frac{1}{N}\sum^{N}_{k=1}x_k-\sqrt[N]{\prod^{N}_{k=1}x_k}\leq\exp\left(\frac{\sum^{N}_{k=1}x_k\log x_k}{\sum^{N}_{k=1}x_k}\right)\tag 1
$$
A: Idea for a proof :
As the inequality is homogenous we obtain an equivalent inequality like :
$$\left(\frac{1}{x}\right)^{\frac{2}{x^2+1}}+\frac{1}{x}-\frac{1}{x^2}\geq1$$
Using the Bernoulli's first approximation and some others terms in the Newton's expansion of $(1+x)^a$ we have ($0< x\leq 1$).:
$$\left(\frac{1}{x}\right)^{\frac{2}{x^{2}+1}}+\frac{1}{x}-\frac{1}{x^{2}}\geq 1+\frac{1}{6}\cdot\left(\frac{2}{x^{2}+1}-2\right)\left(\frac{2}{x^{2}+1}-1\right)\left(\frac{2}{x^{2}+1}\right)\left(\frac{1}{x}-1\right)^{3}+\left(\frac{1}{x}-1\right)\left(\frac{2}{x^{2}+1}\right)+\frac{1}{x}-\frac{1}{x^{2}}+0.5\left(\frac{2}{x^{2}+1}-1\right)\left(\frac{2}{x^{2}+1}\right)\left(\frac{1}{x}-1\right)^{2}\geq 1$$
The RHS becomes (using Wolfram alpha for the simpplification) :
$$ \frac{1}{3}\frac{(3x^6-14x^5+27x^4-28x^3+17x^2-6x+1)}{(x(x^2+1)^3)}\geq 0$$
Or :
$$\frac{1}{3}\frac{(x-1)^4 (3x^2-2x+1)}{(x(x^2+1)^3)}\geq 0$$
Wich is obvious !
Done !
For a reference of proof without calculus of the binomial theorem see :
Reference :
https://www.jstor.org/stable/2319010?seq=1
Binomial theorem proof for rational index without calculus
A: This is not an answer but a remark that I think everyone on this topic has missed so far.( That is why I posted as an answer)
The ‘derivative’ solution presented by OP’s teacher is WRONG.
The function $f(x)$ is not decreasing when $x\le y$ . More precisely, $x=0$ should be another maxima of this function as
$$\lim_{x\rightarrow 0^+}f(x)=0$$
That is also why the ‘solutions’ given by @Yuri and @erik deserve more recognition.
