Difference between harmonic mean of arithmetic means and arithmetic mean of harmonic means

Question

Let $S=\{(x_i, y_i)\}_{i=1...n} \in [0,1]^{2n}$ bet a tuple of ordered pairs, and let $A, H$ denote the arithmetic and harmonic mean. Then $$ \sup_S (H(\underset{i}{A}(x_i),\underset{i}{A}(y_i)) - \underset{i}{A}(H(x_i, y_i))) = \begin{cases} 0.5,n\text{ is even}\\ 0.5 - \frac{1}{2n^2},\text{else} \end{cases} $$ We found a proof (Opitz and Burst, 2019: Macro F1 and Macro F1) by showing that the difference can be increased by intelligently swapping variables and then setting them to either 0 or 1. The proof is fairly long, and we are wondering: Is there a simple way to show this bound?

Answer 1

$ \newcommand{\R}{\mathbb{R}} \newcommand{\la}{\lambda} \newcommand{\p}{\partial} \newcommand{\PP}{\mathcal{P}}$ Let $x:=(x_1,\dots,x_n)\in[0,1]^n$, $y:=(y_1,\dots,y_n)\in[0,1]^n$, $h:=(h_1,\dots,h_n)$, \begin{equation*} h_i:=H(x_i,y_i),\quad H(u,v):=\frac2{\frac1u+\frac1v}=\frac{2uv}{u+v} \end{equation*} for $u>0$ and $v>0$, and, by continuity, $H(u,v):=0$ for $u\ge0$ and $v\ge0$ with $u v=0$. Let $Az:=\frac1n\sum_1^n z_i$ for $z:=(z_1,\dots,z_n)$. Then the result in question can written as \begin{equation*} L:=L(x,y):=H(Ax,Ay)-Ah\le L^*_n:= \left\{ \begin{alignedat}{2} &\frac12&&\text{ if $n$ is even}\\ & \frac12-\frac1{2n^2}&&\text{ if $n$ is odd}, \end{alignedat} \right. \tag{0} \end{equation*} with equality for some $x,y$ in $[0,1]^n$.

The maximum of $L(x,y)$ over all $(x,y)\in[0,1]^n\times[0,1]^n$ is attained. In what follows, let $(x,y)$ be such a maximizer.

With $[n]:=\{1,\dots,n\}$, $p$ and $q$ in $\{0,1\}$, and $|K|:=(\text{cardinality of $K$)}$, let
\begin{gather*} I:=\{i\in[n]\colon 0<x_i<1\},\quad J:=\{i\in[n]\colon 0<y_i<1\},\\ I_p:=\{i\in[n]\colon x_i=p\},\quad J_q:=\{i\in[n]\colon y_i=q\},\\ s_{pq}:=\tfrac1n|I_p\cap J_q|, \end{gather*} so that $s_{00}+s_{01}+s_{10}+s_{11}\le1$.

If $Ax=0$, then $x=0$ and hence $h=0$ and $L=0$, which makes the inequality in (0) trivial. So, without loss of generality (wlog), $Ax>0$. Similarly, wlog $Ay>0$. So, \begin{equation*} r:=Ay/Ax\in(0,\infty). \tag{1} \end{equation*}

Let $\p_u$ denote the partial derivative with respect to a variable $u$. Then
\begin{equation*} \p_u H(u,v)=2\Big(\frac v{u+v}\Big)^2 \end{equation*} for $u>0$ and $v>0$. So, for any $i\in I$ \begin{equation*} \frac n2\,\p_{x_i}L =\Big(\frac r{r+1}\Big)^2-\Big(\frac{y_i}{x_i+y_i}\Big)^2=0, \end{equation*} because $(x,y)$ is a maximizer of $L$. So, $y=rx>0$ on $I$. Similarly, $y=rx>0$ on $J$, and hence $y=rx>0$ on $I\cup J$. So, with $\xi:=\frac1n\,\sum_{i\in I\cup J}x_i$, \begin{alignat*}{5} &Ax=&& &&s_{10}&+&s_{11}&&+\xi, \tag{Ax}\\ &Ay=&&s_{01}&& &+&s_{11}&&+\xi r, \tag{Ay}\\ &Ah=&& && &&s_{11}&&+\xi\frac{2r}{1+r}. \end{alignat*} So, \begin{align*} L&=\frac{2 Ax\,Ay}{Ax+Ay}-Ah \\ &=Ax\frac{2r}{1+r}-\Big(s_{11}+\xi\frac{2r}{1+r}\Big) \\ &=(s_{10}+s_{11})\frac{2r}{1+r}-s_{11}. \tag{2} \end{align*} It also follows from (Ax) and (Ay) that the equality in (1) can be rewritten as \begin{equation*} s_{01}+s_{11}=r(s_{10}+s_{11}). \end{equation*} So, if $s_{10}+s_{11}=0$, then $s_{11}=0$ and hence, by (2), $L=0$. So, wlog $s_{10}+s_{11}>0$ and hence $r=\frac{s_{01}+s_{11}}{s_{10}+s_{11}}$. Using this expression for $r$, we get from (2): \begin{align*} L=M:=\frac{2 s_{01} s_{10} + (s_{01}+ s_{10})s_{11}}{s_{01} + s_{10} + 2 s_{11}}. \end{align*} Next, \begin{equation*} \p_{s_{11}}M:=\frac{(s_{01}-s_{10})^2}{(s_{01} + s_{10} + 2 s_{11})^2}\ge0. \end{equation*} So, wlog one may replace $s_{11}$ by its largest possible value, $1-s_{01}-s_{10}$: \begin{equation*} L=M\le N:=M|_{s_{11}=1-s_{01}-s_{10}}= \frac{(1-s_{01})s_{01}+(1-s_{10})s_{10}}{2-s_{01}- s_{10}}. \end{equation*} Further, \begin{equation*} (\p_{s_{01}}+\p_{s_{10}})N= \frac{4(1-s_{01})(1-s_{10})}{(2-s_{01}-s_{10})^2}\ge0. \end{equation*} So, if we increase $s_{01}$ and $s_{10}$ by the same amount, while keeping $s_{01}+s_{10}\le1$, the value of $N$ may only increase. So, \begin{equation*} L\le N|_{s_{10}=1-s_{01}}=2(1-s_{10})s_{10} \le2(1-\tfrac mn)\tfrac mn=L^*_n, \end{equation*} where $m:=\lfloor n/2\rfloor$; the latter inequality follows because $(1-u)u$ is decreasing in $|u-1/2|$ for $u\in[0,1]$.

The inequality in (0) turns into the equality if $(x_i,y_i)=(1,0)$ for $i=1,\dots,m$ and $(x_i,y_i)=(0,1)$ for $i=m+1,\dots,n$.

The entire proof is now complete.

Answer 2

I will give you a short proof for the even case. I think it may be possible to imitate the steps for $n$ odd.

Restating your problem in a math-olympiad fashion, one has to prove that for $0< x_i,y_i\leq 1$, $1\leq i\leq n$, the following holds:

$$ \frac{1}{n} \cdot \frac{2\left(\sum_{i=1}^n x_i\right) \left(\sum_{i=1}^n y_i\right)}{\left(\sum_{i=1}^n x_i\right) + \left(\sum_{i=1}^n y_i\right)} - \frac{2}{n} \sum_{i=1}^n \frac{x_iy_i}{x_i+y_i} \leq \frac{1}{2}$$

Which in turn, using the inequalities between harmonic and arithmetic means in the first summand, means that it is enough to prove:

$$ \frac{1}{2n} \left(\sum_{i=1}^n x_i + \sum_{i=1}^n y_i\right) - \frac{2}{n} \sum_{i=1}^n \frac{x_iy_i}{x_i+y_i} \leq \frac{1}{2}$$

And this is equivalent to this:

$$ \sum_{i=1}^n (x_i + y_i) - 4 \sum_{i=1}^n \frac{x_iy_i}{x_i+y_i} \leq n$$

Which, rewriting is just to prove:

$$ \sum_{i=1}^n \frac{(x_i-y_i)^2}{x_i+y_i} \leq n$$

And it is sufficient to verify that each summand is $\leq 1$, which is pretty simple given that $(x_i-y_i)^2\leq x_i+y_i$, since this is just equivalent to $x_i^2+y_i^2 \leq x_i+y_i + 2x_iy_i$, and the last inequality follows directly from the fact that $x_i^2\leq x_i$ and $y_i^2\leq y_i$, for $x_i,y_i\in [0,1]$.

Interestingly, the case of equality is straightforward to construct, since we only need that $(x_i-y_i)^2= x_i+y_i$ which is easy to see only can happen for $x_i=1$ and $y_i=0$ or viceversa, and that $\sum x_i = \sum y_i$.