$I_{1}=\frac{E_{1}+E_{2}}{R+r_{1}+r_{2}}$

After short circuiting the battery of $\mathrm{emf}$ $\mathrm{E}_{2}$. current through resistance $\mathrm{R}$ would be,

$I_{2}=\frac{E_{1}}{R+r_{1}}$

Now, $\quad \mathrm{I}_{2}>\mathrm{I}_{1}$

$\therefore $ ${\frac{\mathrm{E}_{1}}{\mathrm{R}+\mathrm{r}_{1}}>\frac{\mathrm{E}_{1}+\mathrm{E}_{2}}{\mathrm{R}+\mathrm{r}_{1}+\mathrm{r}_{2}}} $

${\mathrm{E}_{1} \mathrm{r}_{2}>\mathrm{E}_{2}\left(\mathrm{R}+\mathrm{r}_{1}\right)}$