<s>[INST] 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.
 [/INST] 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
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