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,
$$\begin{align*} DE + DF &= c \cos C + b \cos B \\ &= (2R\sin C) \cos C + (2R\sin B) \cos B \\ &= R(\sin 2C + \sin 2B) \\ &= 2R\sin (B+C) \cos (B-C) \\ &= 2R\sin A \cos (B-C) \\ &= a \cos (B-C) \le a, \end{align*}$$
where $R$ is the radius of the circumcircle of $\triangle ABC$.
That is,
$$DE + DF \le a.$$
Similarly,
$$DE + EF \le b \text{ and } EF + DF \le c.$$
Therefore,
$$DE + DF + EF \le \frac{1}{2}(a+b+c).$$