# 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 ?

• 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. – Connor Harris Jun 6 at 13:52
• This is not new, actually (math.stackexchange.com/questions/1432043/…) – Jack D'Aurizio Jul 29 at 21:39

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.

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 $$\sf{(1)}$$.
• 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. – Morgan Rogers Jun 13 at 7:47