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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.

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  • $\begingroup$ I rewrote the RHS as $\sqrt[x+y]{x^x y^y}$ and showed that this was greater than or equal to $\frac{x+y}{2}$ (briefly: apply weighted AM-GM to $1/x$ and $1/y$ with respective weights $x$ and $y$, and then invert). But this doesn't help, as it reduces proving the original inequality to showing that $\frac{x+y}{2} \leq \sqrt{xy}$, which obviously isn't true. $\endgroup$ Jun 6 '19 at 13:52
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    $\begingroup$ This is not new, actually (math.stackexchange.com/questions/1432043/…) $\endgroup$ Jul 29 '19 at 21:39
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    $\begingroup$ @JackD'Aurizio: The two inequalities do not look the same, though. $\endgroup$
    – Hans
    Nov 5 '19 at 23:30
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    $\begingroup$ This problem is equivalent to showing that given Riesz conjugates $p$ and $q$ (i.e. $1/p+1/q=1$) then the inequality below holds $$\sqrt{pq}+p^{1/p}q^{1/q}\geq p^{1/2+1/p}q^{1/2+1/q}$$ Maybe someone can get Holder's Inequality to make it work. $\endgroup$
    – user123641
    Nov 13 '19 at 3:47
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    $\begingroup$ @JackD'Aurizio : math.stackexchange.com/questions/1432043/… is too weak $\endgroup$
    – user90369
    Nov 14 '19 at 10:35
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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:

enter image description here

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

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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.

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  • $\begingroup$ Since we are working with positive values it might be conceptually easier to substitute $a=b^2$. I tried expanding the exponential, which is acceptable since it's one of the many definitions of exp, but no finite number of terms seems to give the bound for all $x$, which suggests that any method not involving differentiation is likely to fail. $\endgroup$ Jun 13 '19 at 7:47
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A COMMENT.

Actualy holds the following (somewhat) conjecture generalization

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

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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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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.

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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.

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  • $\begingroup$ Using a Maclaurin Series is just a more complicated way of using derivatives. The OP already knows you can show that using calculus on the derivative you can get the result. $\endgroup$
    – Eric
    May 22 at 3:42
  • $\begingroup$ @Eric You are wrong. However, thanks for the comment - note about binomial series added. $\endgroup$ May 22 at 5:03

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