High School Geometry - If $BC$ is the greatest side of $\triangle ABC$, $D$ & $E$ are points on $BC, CA$... 
If $BC$ is the greatest side of $\triangle ABC$, and $D$ & $E$ are points on $BC$ & $CA$, respectively, prove that $BC \ge DE$.

Clearly, equality holds iff $D$ is on $B$ and $E$ is on $C$. 
I have tried to prove the inequality comparing $\angle BAC$ and $\angle DEC$.
If $\angle DEC \ge \angle BAC,$ then $DC$ hence $BC$ is $\gt DE$.
Else, on similar grounds $DE\lt EC$ hence $\lt BC$.
Is there something I am missing out. And, most importantly, can someone suggest a more neat version?
 A: We assume as known that in any triangle $PQR$, the largest side is opposite the largest angle, and conversely. 
Lemma: Let $PQR$ be a triangle, and let $X$ be in the interval $PR$, and not at an endpoint. Then $QX \lt QP$ or $QX \lt QR$ (or both). More compactly, $QX\lt \max(QP,QR)$. 
Proof: Since angles $QXP$ and $QXR$ add up to $180^\circ$, one at least is $\ge 90^\circ$. So $\angle QXP$ is the largest angle in $\triangle QXP$, or $\angle QXR$ is the largest angle in $\triangle QXR$, or both. If $\angle QXP$ is the largest angle in $\triangle QXP$, then $QP \gt QX$. Similarly, if $\angle QXR$ is the largest angle in $\triangle QXR$, then $QR\gt QX$.
Now we look at our triangle $ABC$.  Suppose that $D=B$. If $E=C$, then $DE=BE=BC$. If $E\ne C$, then by the Lemma $BE$ (that is, $DE$) is $\lt\max(BC,BA)=BC$.  
Suppose now that $D\ne B$. The case $E=C$ is easy. 
If $B\ne D$ and $E\ne C$, draw the line $BE$. By the Lemma, we have $BE\lt BC$. Again by the Lemma, $DE\lt \max(BE,CE)\lt BC$. This completes the proof.   
A: 
Here is the answer in picture hope it is right
It works for each case and for d lying on any vertice we can prove by exterior angle theorem easily
A: I have a geometric solution:
By extending $ED$ and $AB$ will intersect at $F$, assuming they are not parallel. Thus in $D$ $\triangle AEF$ we observe $\angle FAE=\angle BAC<\angle DEC$, and since $\angle BAC$ is the greatest angle we have $DC>DE$.
