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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. $\endgroup$ – Connor Harris Jun 6 at 13:52
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    $\begingroup$ This is not new, actually (math.stackexchange.com/questions/1432043/…) $\endgroup$ – Jack D'Aurizio Jul 29 at 21:39
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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 $\sf{y=ax}$ for some $\sf{a,x>0}$. Then \begin{align}\sf{x+y-\sqrt{xy}\leq\exp\left(\frac{x\ln x+y\ln y}{x+y}\right)}&\impliedby\sf{x+ax-x\sqrt a\le\exp\left(\frac{x\ln x+ax\ln ax}{x+ax}\right)}\\&\impliedby\sf{x(1-\sqrt a+a)\le\exp\left(\ln x+\frac{a\ln a}{1+a}\right)}\\&\impliedby\sf{1-\sqrt a+a\le a^{\frac a{1+a}}}\end{align} so it suffices to show that $\sf{(1-\sqrt a+a)^{a+1}\le a^a}.\tag1$

It may be worth noting that the inequality is extremely tight which can be seen via this visualisation, and the Bernoulli inequality for $\sf{(1-\sqrt a+a)^{a+1}\ge1+(a+1)(a-\sqrt a)}$ is too weak to prove $\sf{(1)}$.

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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. $\endgroup$ – Morgan Rogers Jun 13 at 7:47

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