China Girls' Mathematical Olympiad

Proof Since $\angle ADB=\angle AEB=90^{\circ }$, so four points $A$, $B$, $D$ and $E$ are concyclic, and furthermore, $AB$ is the diameter. Hence, by the sine rule, we can get $$\frac{DE}{\sin \angle DAE}=AB=c,$$ so $$DE=c\sin \angle DAE.$$ In addition, $\angle DAC+\angle DCA=90^{\circ }$, therefore, $$DE=c\cos C.$$ Similarly, we can get $DF=b\cos B$. Therefore, $$\begin{array}{rl}DE+DF& =c\cos C+b\cos B\\ & =(2R\sin C)\cos C+(2R\sin B)\cos B\\ & =R(\sin 2C+\sin 2B)\\ & =2R\sin (B+C)\cos (B-C)\\ & =2R\sin A\cos (B-C)\\ & =a\cos (B-C)\le a,\end{array}$$ where $R$ is the radius of the circumcircle of $△ABC$. That is, $$DE+DF\le a.$$ Similarly, $$DE+EF\le b\text{ and }EF+DF\le c.$$ Therefore, $$DE+DF+EF\le \frac{1}{2}(a+b+c).$$