In A B C, ∠ B < ∠ C . If D and E are...

Geometry of a triangle is studied extensively. Let the sides be $a$,$b$,$c$ respectively and $l_a$ be the length of angle bisector from vertex $A$ onto the side whose length is $a.$ The Angle Bisector property tells us that $l_a$ bisects the side $BC$ in the proportion $AC:AB=b:c.$ This gives us two equations:
$\{\begin{array}{l}x+y=a\\ {\displaystyle \frac{x}{y}}={\displaystyle \frac{b}{c}}\end{array}$
Solving this is easy and it yields:
$x={\displaystyle \frac{ab}{b+c}},y={\displaystyle \frac{ac}{b+c}}.$
Finally, Stewart's theorem gives us:
$a\cdot l_a^2={\displaystyle \frac{b^{2}ac+c^{2}ab}{b+c}}-{\displaystyle \frac{a^{3}bc}{(b+c{)}^2}}$
or
$l_a=\sqrt{bc-{\displaystyle \frac{a^{2}bc}{(b+c{)}^2}}}={\displaystyle \frac{2\sqrt{bcp(p-a)}}{b+c}}$
where $p={\displaystyle \frac{a+b+c}{2}}$ is the semiperimeter.

With this formula in hand and in the setup of your problem, one can immediately see that:
$C{E}^2-B{D}^2=l_c^2-l_b^2=ab-{\displaystyle \frac{c^{2}ab}{(a+b{)}^2}}-ac+{\displaystyle \frac{b^{2}ac}{(a+c{)}^2}}=$
$=a(b-c)(1+{\displaystyle \frac{bc(a^2+b^2+c^2+bc+2ab+2ac)}{(a+b{)}^2(a+c{)}^2}})\le 0.$
Thus, $l_b\ge l_c.$