Inequality concerning sums of ordered variables With $Y=(Y_1,\ldots,Y_{n_y})$, $n_y \geq 2$,  and $Z=(Z_1,\ldots,Z_{n_z})$, $n_z \geq 2$, writing 
$Y_{(1)},Y_{(2)},..., Y_{(n_y)}$ the ordered values of $Y$ such that: $Y_{(1)} < Y_{(2)} < ... < Y_{(n)}$ (and the same for $Z_{(i)}$), assuming only positive values for $Y$ and $Z$, and setting the merged vector (concatenated) $(Y^\frown Z)=(Y_1,\ldots,Y_{n_y},Z_1,\ldots,Z_{n_z})$,
trying to prove the following inequality:
$$\frac{ \sum _{i=1}^{n_y+n_z} i (Y^\frown Z)_{(i)}}{\sum _{i=1}^{n_y} Y_i+\sum _{i=1}^{n_z} Z_i}-\frac{ \sum _{i=1}^{n_y} i Y_{(i)}}{\sum _{i=1}^{n_y} Y_i}-\frac{\sum _{i=1}^{n_z} i Z_{(i)}}{\sum _{i=1}^{n_z} Z_i}+\frac{1}{2}\geq 0$$
$\textbf{Approach: }$ 
Using the rearrangement inequality (permutation group), we can derive
$$\sum _{k=1}^{n_z+n_y} k (Y^\frown Z)_{(k)}\geq \sum _{j=i}^{n_y} (j+n_z) Y_{(j)}+\sum _{i=1}^{n_z} i Z_{(i)}=n_z \sum _{j=i}^{n_y} Y_{j}+\sum _{j=i}^{n_y} j Y_{(j)}+\sum _{i=1}^{n_z} i Z_{(i)}$$
$$ \sum _{k=1}^{n_z+n_y} k (Y^\frown Z)_{(k)}\geq \sum _{i=1}^{n_y} \text{iY}_{(i)}+\sum _{j=i}^{n_z} (j+n_y) Z_{(j)}=n_y \sum _{i=i}^{n_z} Z_i+\sum _{j=i}^{n_y} j Y_{(j)}+\sum _{i=1}^{n_z} i Z_{(i)}$$
I wonder if this route leads anywhere.
 A: We can show this using induction.
Step 1 (base case)- The inequality is correct for $n_y=n_z=1$. Without loss of generality, assume $Z\geq Y$. In this case, $S_{s}=Y+2Z$, $S_{s1}=S_1=Y$, $S_{s2}=S_2=Z$. Therefore, the left hand side of the inequality becomes
$$
LHS=\frac{Y+2Z}{Y+Z}-\frac{Y}{Y}-\frac{Z}{Z}+\frac{1}{2}=\frac{Z}{Y+Z}-\frac{1}{2}\geq 0.
$$
Step 2 (inductive step)- Assuming that the inequality is correct for any vectors Y and Z of length $n_y=k_y$ and $n_z=k_z$, implies that for $n_y=k_y+1$ and $n_z=k_z$, the inequality is correct.
Proof of step 2:
Define $Y’$ of length $k_y$ such that $Y’_{(k_y)}$=$Y_{(k_y)}+\frac{k_y+1}{k_y}Y_{(k_y+1)}$ and the rest of the entries are the same as $Y$. Note by this definition, we get:
$$
S’_1=S_1+\frac{1}{k_y}Y_{(k_y+1)}\geq S_1,\\
S’_{s1}=\sum_{i=1}^{k_y}iY'_{(i)}=k_y(Y_{(k_y)}+\frac{k_y+1}{k_y}Y_{(k_y+1)})+\sum_{i=1}^{k_y-1}iY_{(i)}=\sum_{i=1}^{k_y+1}iY_{(i)}=S_{s1},\\
S'_s\leq S_s.
$$
According to our assumption for vectors of length $n_y=k_y$ and $n_z=k_z$, for $Y’$ and $Z$ we have
$$
\frac{S’_s}{S’_1+S_2}-\frac{S’_{s1}}{S’_1}-\frac{S_{s2}}{S_2}+\frac{1}{2} \geq 0.
$$
Since $S’_1\geq S_1$, $S’_{s1}=S_{s1}$, and $S'_s\leq S_s$,
we get 
$$
0\leq\frac{S’_s}{S'_1+S_2}-\frac{S'_{s1}}{S'_1}-\frac{S_{s2}}{S_2}+\frac{1}{2}\\
=\frac{S’_s}{S'_1+S_2}-\frac{S_{s1}}{S'_1}-\frac{S_{s2}}{S_2}+\frac{1}{2}\\
\leq\frac{S’_s}{S_1+S_2}-\frac{S_{s1}}{S_1}-\frac{S_{s2}}{S_2}+\frac{1}{2}\\
\leq\frac{S_s}{S_1+S_2}-\frac{S_{s1}}{S_1}-\frac{S_{s2}}{S_2}+\frac{1}{2}.
$$
Same procedure can be repeated for $Z$ vector. Hence, by induction, the inequality is proved.
