Let points $D$ and $D'$ are symmetrical to the points $A$ and $A'$ respectively to the $BC$ . Let draw a bisector $AK$ and $A_1K_1$ of our triangles.Notice that the points $K$ and $K'$ are centres of circles inscribed in quadrilaterals $ABDC$ and $A_1B_1D_1C_1$ required inequality turned to the obvious inequality $r$ > $r_1$ where $r$ and $r_1$ are radiis of indicated circles

Let the sides of $\triangle ABC$ and $\triangle A_1B_1C_1$ are $a,b,c,a_1,b_1,c_1$, their radius and altitudes to the side $\overline{BC}$ are $r,r_1,h,h_1$ It's known that $r>r_1$. So, $\frac{S_1}{a_1+b_1+c_1}<\frac{S}{a+b+c}$ , Combine this by $h>h_1\Longrightarrow S_1a<Sa_1$ We get $\frac{S_1}{b_1+c_1}<\frac{S}{b+c}$ as desired