Proving a geometric inequality $BE+CF>EF$ In the figure $D$ is the midpoint of $BC$ and $\Delta EDF$ is right triangle.Prove that $BE+CF>EF$

I tried triangle inequality $$BE+ED>BD$$
$$CD+FC>DF$$
Add:$$BE+FC>DF-ED$$
I am stuck now!


Rotate the figure by $180^\circ$ around $D$.
Now the desired inequality becomes $BE + BF' > EF$.
However $EF = EF'$ by congruent triangles $\triangle EDF' \cong \triangle EDF$.
Finish off using triangle inequality.
