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?

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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.

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"The largest side is opposite the largest angle, and conversely." I call the broader principle the Three Bears Theorem: The Poppa-Bear side is opposite Poppa-Bear angle; the Momma-Bear side is opposite Momma-Bear angle; and the Baby-Bear side is opposite Baby-Bear angle. :) – Blue Jan 2 '13 at 19:02
Aren't we supposed to mention about the case when neither point is an endpoint ? – Aseem Dua Jan 2 '13 at 19:35
Of course, that's what the post does. That's taken care of by the last two sentences, I will separate them out so it will be more clear. – André Nicolas Jan 2 '13 at 20:41
Yeah, got mixed up with names there.. Nice solution. – Aseem Dua Jan 3 '13 at 2:37