We sampled $n$ observations $X_1,...,X_n$ from a location-scale family's distribution and $Y_1,...,Y_m$ from the same distribution but different parameters.
How does one show that
$\frac{\sum_{1}^{n}\left(X_{i}-\bar{X}\right)^{2}+\sum_{1}^{m}\left(Y_{i}-\bar{Y}\right)^{2}}{\sum_{1}^{n}\left\{X_{i}-[(n \bar{X}+m \bar{Y}) /(n+m)]\right\}^{2}+\sum_{1}^{n}\left\{Y_{i}-[(n \bar{X}+m \bar{Y}) /(n+m)]\right\}^{2}} =\frac{\sum_{1}^{n}\left(X_{i}-\bar{X}\right)^{2}+\sum_{1}^{m}\left(Y_{i}-\bar{Y}\right)^{2}}{\sum_{1}^{n}\left(X_{i}-\bar{X}\right)^{2}+\sum^{m}\left(Y_{i}-\bar{Y}\right)^{2}+[n m /(n+m)](\bar{X}-\bar{Y})^{2}}?$
Assume these are true
$$ \sum_{1}^{n}\left(X_{i}-\frac{n \bar{X}+m \bar{Y}}{n+m}\right)^{2}=\sum_{1}^{n}\left[\left(X_{i}-\bar{X}\right)+\left(\bar{X}-\frac{n \bar{X}+m \bar{Y}}{n+m}\right)\right]^{2} $$ $$ =\sum_{1}^{n}\left(X_{i}-\bar{X}\right)^{2}+n\left(\bar{X}-\frac{n \bar{X}+m \bar{Y}}{n+m}\right)^{2} $$ and $$ \begin{aligned} \sum_{1}^{m}\left(Y_{i}-\frac{n \bar{X}+m \bar{Y}}{n+m}\right)^{2} &=\sum_{1}^{m}\left[\left(Y_{i}-\bar{Y}\right)+\left(\bar{Y}-\frac{n \bar{X}+m \bar{Y}}{n+m}\right)\right]^{2} \\ &=\sum_{1}^{m}\left(Y_{i}-\bar{Y}\right)^{2}+m\left(\bar{Y}-\frac{n \bar{X}+m \bar{Y}}{n+m}\right)^{2} . \end{aligned} $$ But $$ n\left(\bar{X}-\frac{n \bar{X}+m \bar{Y}}{n+m}\right)^{2}=\frac{m^{2} n}{(n+m)^{2}}(\bar{X}-\bar{Y})^{2} $$ and $$ m\left(\bar{Y}-\frac{n \bar{X}+m \bar{Y}}{n+m}\right)^{2}=\frac{n^{2} m}{(n+m)^{2}}(\bar{X}-\bar{Y})^{2} $$
$$\frac{m^2 n}{(n+m)^2} + \frac{n^2 m}{(n+m)^2} = \frac{nm}{(n+m)^2} (m+n) = \frac{nm}{n+m}.$$