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?
2 Answers
$ \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}:=\tfrac1nI_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, $1s_{01}s_{10}$:
\begin{equation*}
L=M\le N:=M_{s_{11}=1s_{01}s_{10}}=
\frac{(1s_{01})s_{01}+(1s_{10})s_{10}}{2s_{01} s_{10}}.
\end{equation*}
Further,
\begin{equation*}
(\p_{s_{01}}+\p_{s_{10}})N=
\frac{4(1s_{01})(1s_{10})}{(2s_{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}=1s_{01}}=2(1s_{10})s_{10}
\le2(1\tfrac mn)\tfrac mn=L^*_n,
\end{equation*}
where $m:=\lfloor n/2\rfloor$; the latter inequality follows because $(1u)u$ is decreasing in $u1/2$ 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.

$\begingroup$ I have added a couple of details to the proof, and also fixed a couple of typos. $\endgroup$ Commented Nov 27, 2019 at 4:20
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 matholympiad 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:
$$ \sum_{i=1}^n \frac{(x_iy_i)^2}{x_i+y_i} \leq n$$
And it is sufficient to verify that each summand is $\leq 1$, which is pretty simple given that $(x_iy_i)^2\leq x_i+y_i$, 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 $(x_iy_i)^2= x_i+y_i$ which is easy to see only can happen for $x_i=1$ and $y_i=0$ or viceversa, and that $\sum x_i = \sum y_i$.

$\begingroup$ Interesting! It would probably have never occurred to me that simply bounding the harmonic mean (of the arithmetic means) by the corresponding arithmetic mean would work, even for even $n$. However, after that, you are only rewriting the bound on the difference, and I don't see how this could be made work for odd $n$. $\endgroup$ Commented Nov 27, 2019 at 22:58