$(I,r),(I_a,r_a)$ are the incircle and A-excircle of ABC. From $\triangle ABC \sim \triangle ADE$ with A-excircles $(I_a,r_a),(I,r),$ we get $\frac{DE}{BC}=\frac{r}{r_a} \Longrightarrow \ DE=\frac{a \cdot r}{r_a}.$ Hence, we must have $\frac{a+b+c}{8} \ge \frac{a \cdot r}{r_a}$ Using the identity $\frac{r}{r_a} =\frac{b+c-a}{a+b+c},$ the desired inequality becomes: $\frac{(a+b+c)^2}{8}-a(b+c-a) \ge 0 \Longleftrightarrow \frac{(b+c-3a)^2}{8} \ge 0,$ which is true Equality holds for $\triangle ABC$ with $b+c=3a,$ i.e. when $DE \cap (I)$ is the Nagel point

I think I have easier solution. $BCED$ is tangent quadrilateral $\implies DE=DB+EC-DC \leftrightarrow DE=\dfrac{ac+ab-a^{2}}{a+b+c}$ So, $\dfrac{ac+ab-a^{2}}{a+b+c}\leq\dfrac{1}{8}(a+b+c) \leftrightarrow (b+c-3a)^2\geq 0$, and we are finished.

TheBernuli wrote: I think I have easier solution. $BCED$ is tangent quadrilateral $\implies DE=DB+EC-[color=\#FF0000][b]BC[/b][/color] \leftrightarrow DE=\dfrac{ac+ab-a^{2}}{a+b+c}$ So, $\dfrac{ac+ab-a^{2}}{a+b+c}\leq\dfrac{1}{8}(a+b+c) \leftrightarrow (b+c-3a)^2\geq 0$, and we are finished. I did not get how does that come 'easier'! Do you have any secret formula to calculate $BD,CE$? Didn't you use the same $\frac{AD}{AB}=\frac{AE}{AC}=\frac{r}{r_a}$, thus rendering the calculations heavier?? Best regards, sunken rock

Dear Mathlinkers, I research a more geometrical proof... Any ideas? Sincerely Jean-Louis

Dear Mathlinkers, a proof using areas is possible... next... Sincerely Jean-Louis