New bound for Am-Gm of 2 variables Today I'm interested by the following problem :


  Let $x,y&gt;0$ 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.
The alternative forms of the given inequality are
$$x\ln x + y \ln y \ge (x+y)\ln(x+y)+(x+y)\ln\left(1-\dfrac{\sqrt{xy}}{x+y}\right),$$
$$x\ln\,\dfrac x{x+y} + y \ln\, \dfrac y{x+y} \ge (x+y)\ln\left(1-\dfrac{\sqrt{xy}}{x+y}\right),$$
$$\dfrac x{x+y}\,\ln\,\dfrac x{x+y} + \dfrac y{x+y}\, \ln\, \dfrac y{x+y} 
\ge \ln\left(1-\sqrt{\dfrac x{x+y}}\;\sqrt{\dfrac y{x+y}}\right).\tag1$$
Let WLOG $\;x\le y,\;$
$$\dfrac yx = \dfrac{1+z}{1-z},\quad \dfrac x{x+y} =\dfrac{1-z}{2}
,\quad \dfrac y{x+y} =\dfrac{1+z}{2},\quad z\in[0,1],\tag2$$
then $(1)$ can be presented in the forms of
$$\dfrac{1-z}{2}\,\ln\dfrac{1-z}{2}+\dfrac{1+z}{2}\,\ln\dfrac{1+z}{2} \ge \ln(1-\,^1\!/_2\sqrt{1-z^2}\,),$$
$$\dfrac{1-z}{2}\,\ln(1-z)+\dfrac{1+z}{2}\,\ln(1+z) \ge \ln(2-\sqrt{1-z^2}),$$
$$-\dfrac{z}{2}\,\ln(1-z)+\dfrac{z}{2}\,\ln(1+z) \ge \ln(2-\sqrt{1-z^2})
-\ln\sqrt{1-z^2},$$
$$f(z) = (1-z)^{-\large\,^z\!/_2}(1+z)^{\large\,^z\!/_2}+1- \dfrac 2{\sqrt{1-z^2}}\ge 0. \tag3$$
Using binomial series in the forms of
$$(1-z)^{-\large\,^z\!/_2} = 1 + \dfrac12 z^2 + \dfrac1{2!}\,\dfrac z2\,\dfrac{2+z}2 z^2
 + \dfrac1{3!}\,\dfrac z2\,\dfrac{2+z}2\dfrac{4+z}2 z^3 +\dots$$
$$= 1+\dfrac12z^2+\dfrac14z^3+\dfrac7{24}z^4+\dfrac14z^5+\dfrac{113}{480}z^6+\dots,$$
$$(1+z)^{\large\,^z\!/_2} = 1 + \dfrac12 z^2 + \dfrac1{2!}\,\dfrac z2\,\dfrac{-2+z}2 z^2
 + \dfrac1{3!}\,\dfrac z2\,\dfrac{-2+z}2\dfrac{-4+z}2 z^3 +\dots$$
$$= 1+\dfrac12z^2-\dfrac14z^3+\dfrac7{24}z^4-\dfrac14z^5+\dfrac{113}{480}z^6+\dots,$$
$$(1-z^2)^{-\large\,^1\!/_2} = 1+\dfrac12z^2+\dfrac38z^4+\dfrac5{16}z^6+\dots,$$
one can get the Maclaurin series in the form of
$$f(z) = \left(1+\dfrac12z^2+\dfrac7{24}z^4+\dfrac{113}{480}z^6+\dots\right)^2
-\left(\dfrac14z^3+\dfrac14z^5+\dots\right)^2$$
$$+1-2\left(1+\dfrac12z^2+\dfrac38z^4+\dfrac5{16}z^6+\dots\right),$$
$$f(z) = \dfrac1{12}z^4+\dfrac3{40}z^6+\dots$$
(see also WA series),
with the all positive coefficients.
Proved!
$\color{brown}{\textbf{About binomial series.}}$

Newton's binomial formulas exist independently of the derivatives associated theory.

If the exponent is the rational, then the coeffitients of the binomial series $(1+x)^{\large\frac pq} = 1+a_1x+a_2x^2+\dots$ are defined via the identity
$$(1+a_1x+a_2x^2+\dots)^q = (1+x)^q,$$
by the comparation of the polynomials in the LHS and RHS.

If the exponent is real, then the binomial series are defined via the limiting transition.


These elementary knowledges point that the binomial series exist independently of the derivatives associated theory.