# Simplify to terms of generalized power mean

I have the following expression${}^1$ I'd like to simplify $$r \equiv \left[ \frac1{x\sqrt{x}} \left( 1 - \frac{k}{x} \right) - \frac1{y\sqrt{y}} \left( 1 - \frac{k}{y} \right) \right]\left[ \frac1{\sqrt{x}} \left( 1 - \frac{k}{x} \right) - \frac1{\sqrt{y}} \left( 1 - \frac{k}{y} \right)\right]^{-1}$$ where $x > y > k\geq 1$.

Some attempts led me to think that $r$ can be expressed nicely in terms of generalized means of $x,y$ like the harmonic ($p=-1$), geometric ($p \to 0$), and other powers:${}^2$ \begin{align} H &\equiv \left( \frac1x + \frac1y\right)^{-1} & G &\equiv \sqrt{xy} & M \equiv \left( x^p + y^p \right)^{1/p} \quad \text{with perhaps} \quad p=\frac{-1}2 \end{align}

To be specific, my question is this:

Is there a way to write it in this form $$r \overset{?}{=} \frac1x + \frac1y + \frac1{\sqrt{xy}} + {}\color{red}{??} =\frac1H + \frac1G + {}\color{red}{??}$$ such that the $\color{red}{??}$ part is "nice" in terms of $k$ and maybe $H,G$ or $M$ defined above?

I also wonder if there's some standard techniques like examining the asymptotic form to guess the coefficients of the proposed terms. I tried a bit but didn't get very far.

Directly pulling out $\frac1H$ and $\frac1G$ just change $r$ into something equally inviting but not more compact.

Any suggestions will be appreciated. Honestly, one main reason for me to think that "oh there must be some nice and more compact forms" is the glaring symmetry. I will be okay if there turn out to be none. Thanks.

Footnote 1: $\quad$For the record, $r$ is part of the expression of the ratio between $\frac{\partial^2 f(u)}{ \partial u^2}$ and $\frac{\partial f(u) }{ \partial u}$, evaluated at $u=0$, where $$f(u) = \left( \frac1{\sqrt{u+y}} - \frac1{\sqrt{u+x}} \right)\sqrt{u+k}$$

Footnote 2: $\quad$This is a slight abuse of notations, as the generalized means should carry the averaging factors $\frac12$ or $\frac1{\sqrt{2}}$. My shorthands can also match the p-norms, which has the benefit of the norms ordered in powers have correspondingly decreasing magnitudes. However, p-norms doesn't cover $\sqrt{x y}$ like generalized mean.

## Update (Nov.28th, 2017)

With all the inverses lying around, it turns out that the best way is to express $r$ in terms of the ... you guessed it ... inverses: $v \equiv 1/x$ and $w \equiv 1/y$. I'll spare the readers of the actual algebra. Suffice to say that this is also the natural course to take to adopt the notions of p-norms; one just have to give up on unifying $1 / \sqrt{xy} = 1/G = \sqrt{vw}$ formally.

We can write $$r=\frac 1H+\frac 1G+\color{red}{\frac{1}{M-G+\frac{GM}{k}}}$$ where $$M=(x^{-1/2}+y^{-1/2})^{-2}$$

If we write $$r=\frac{\frac{1}{x\sqrt x}-\frac{1}{y\sqrt y}-k\left(\frac{1}{x^2\sqrt x}-\frac{1}{y^2\sqrt y}\right)}{\frac{1}{\sqrt x}-\frac{1}{\sqrt y}-k\left(\frac{1}{x\sqrt x}-\frac{1}{y\sqrt y}\right)}=\frac 1x+\frac 1y+\frac{1}{\sqrt{xy}}+A$$ then we have $$A=\frac{k\left(\frac 1x-\frac 1y\right)\left(\frac{1}{x\sqrt y}+\frac{1}{y\sqrt x}\right)}{\frac{1}{\sqrt x}-\frac{1}{\sqrt y}-k\left(\frac{1}{x\sqrt x}-\frac{1}{y\sqrt y}\right)}$$

Using $$x+y=\frac{G^2}{H},\quad xy=G^2,\quad \frac{1}{M}=\frac 1H+\frac 2G$$ we have $$\left(\frac 1x-\frac 1y\right)^2=\left(\frac 1H+\frac 2G\right)\left(\frac 1H-\frac 2G\right)\implies \frac 1x-\frac 1y=-\sqrt{\frac 1M\left(\frac 1H-\frac 2G\right)}$$

$$\left(\frac{1}{x\sqrt y}+\frac{1}{y\sqrt x}\right)^2=\frac{1}{G^2}\left(\frac 1H+\frac 2G\right)\implies \frac{1}{x\sqrt y}+\frac{1}{y\sqrt x}=\frac 1G\sqrt{\frac 1M}$$

$$\left(\frac{1}{\sqrt x}-\frac{1}{\sqrt y}\right)^2=\frac 1H-\frac{2}{G}\implies \frac{1}{\sqrt x}-\frac{1}{\sqrt y}=-\sqrt{\frac 1H-\frac{2}{G}}$$

$$\small\left(\frac{1}{x\sqrt x}-\frac{1}{y\sqrt y}\right)^2=\left(\frac 1H+\frac 1G\right)^2\left(\frac 1H-\frac 2G\right)\implies \frac{1}{x\sqrt x}-\frac{1}{y\sqrt y}=-\left(\frac 1M-\frac 1G\right)\sqrt{\frac 1H-\frac 2G}$$

So, we get

$$A=\frac{k\left(-\sqrt{\frac 1M\left(\frac 1H-\frac 2G\right)}\right)\frac 1G\sqrt{\frac 1M}}{-\sqrt{\frac 1H-\frac{2}{G}}-k\left(-\left(\frac 1M-\frac 1G\right)\sqrt{\frac 1H-\frac 2G}\right)}=\frac{1}{M-G+\frac{GM}{k}}$$

The solution by @mathlove is clear and readily generalized to more complicated tasks, which is one of the major things that what I wanted for this.

For the sake of completeness, let me share my original approach that I was dissatisfied with (which prompted the post).

Each step of manipulation is motivated to gain just a near-sighted progress. This route " feels natural" but contains redundancy.

First observe that $r$ takes the form of a "mixture" that is not entirely $\frac1x$ nor entirely $\frac1y$. $$r = \frac{ \frac1x \alpha - \frac1y \beta}{ \alpha - \beta } \quad \text{with} \quad \alpha\equiv \frac1{\sqrt{x}}\left( 1 - \frac{k}{x} \right),~\beta\equiv \frac1{\sqrt{y}}\left( 1 - \frac{k}{y} \right)$$

Here we break the mixture by pulling out a mixture. \begin{align} r &= \frac1{\alpha - \beta}\left\{ \left[ \frac12 \left( \frac1x + \frac1y \right) + \frac12 \left( \frac1x - \frac1y \right) \right] \alpha - \left[ \frac12 \left( \frac1x + \frac1y \right) - \frac12 \left( \frac1x - \frac1y \right) \right] \beta \right\} \\ &= \frac12 \left( \frac1x + \frac1y \right) + \frac{\alpha + \beta}{\alpha - \beta}\frac12 \left( \frac1x - \frac1y \right) \tag{Eq.1} \end{align} The definitions of $\alpha+\beta$ and $\alpha-\beta$ move the product forward naturally. \begin{align} \alpha+\beta &= \frac1{\sqrt{x}} + \frac1{\sqrt{y}} - k \left( \frac1{x^{3/2}} + \frac1{x^{3/2}} \right) = \left( \frac1{\sqrt{x}} + \frac1{\sqrt{y}} \right) \left[ 1 - k \left( \frac1x - \frac1{\sqrt{xy}} + \frac1y \right) \right ] \\ \alpha - \beta &= \frac1{\sqrt{x}} - \frac1{\sqrt{y}} - k \left( \frac1{x^{3/2}} - \frac1{x^{3/2}} \right) = \left( \frac1{\sqrt{x}} - \frac1{\sqrt{y}} \right) \left[ 1 - k \left( \frac1x + \frac1{\sqrt{xy}} + \frac1y \right) \right]. \end{align} \begin{align} \text{so that} \quad &\phantom{{}={}}\frac{\alpha + \beta}{\alpha - \beta}\frac12 \left( \frac1x - \frac1y \right) \\ &= \frac{\alpha + \beta}{\alpha - \beta} \frac12 \left( \frac1{\sqrt{x}} + \frac1{\sqrt{y}} \right) \left( \frac1{\sqrt{x}} - \frac1{\sqrt{y}} \right) \\ &= \frac12 \left( \frac1{\sqrt{x}} + \frac1{\sqrt{y}} \right)^2 \left\{ 1 + \frac{2k}{\sqrt{xy}} \left[ 1 - k \left( \frac1x + \frac1{\sqrt{xy}} + \frac1y \right) \right]^{-1} \right\} \\ &= \left[ \frac12 \left( \frac1x + \frac1y \right) + \frac1{\sqrt{xy}} \right] \left\{ 1 + \frac{2k}{\sqrt{xy}} \left[ 1 - k \left( \frac1x + \frac1{\sqrt{xy}} + \frac1y \right) \right]^{-1} \right\} \end{align}

Substitute this back to Eq.1 (here the redundancy shows) you get $$\begin{gather} r = \frac1x + \frac1y + \frac1{\sqrt{xy}} + A \qquad \text{same as @mathlove opening line} \\ \text{where} \qquad A = \left[ \frac1x + \frac1y + \frac2{\sqrt{xy}} \right] \frac{k}{\sqrt{xy}} \left[ 1 - k \left( \frac1x + \frac1{\sqrt{xy}} + \frac1y \right) \right]^{-1} \end{gather}$$

So this is going through a lot of trouble just to pull the inverses out sequentially. The only difference is that the remaining term $A$ written in this form is readily recognized as

$$A = \frac1M \frac{k}G \left[ 1 - k\left( \frac1H + \frac1G \right) \right]^{-1} \quad \text{or} \quad \frac1M \frac{k}G \left[ 1 - k\left( \frac1M - \frac1G \right) \right]^{-1}$$

depending on your taste.