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 $$ \frac{1}{N}\sum^{N}_{k=1}x_k-\sqrt[N]{\prod^{N}_{k=1}x_k}\leq\exp\left(\frac{\sum^{N}_{k=1}x_k\log x_k}{\sum^{N}_{k=1}x_k}\right)\tag 1 $$
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 :
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.
