Matematika wrote: On sides $ AB$ and $ AC$ of triangle $ ABC$ there are given points $ D,E$ such that $ DE$ is tangent of circle inscribed in triangle $ ABC$ and $ DE \parallel BC$. Prove $ AB + BC + CA\geq 8DE$ Quite easy as a TST. \[ \frac{DE}{a}=\frac{h_a-2r}{h_a}\iff DE=a-\frac{2ar}{h_a}=a-\frac{2a^2sr}{ash_a}=a-\frac{2a^2\Delta}{2\Delta s} =a-\frac{a^2}{s}\] We need to prove that, \[ 2s \geq 8a-\frac{8a^2}{s} \iff 4a^2-4as+s^2 \geq 0 \iff (2a-s)^2 \geq 0\] Done!

Wow! Now I find that it is Italy 1999.

Don't make mistake this isn't IMO TST this is MEMO tst.

Hmm...I know. BTW how did Adrian do?

He didn't write he's IMO contestant

Let $r$ and $r_a$ denote the inradius and $A-$exradius, respectively. It is standard to see that $rs=[ABC]=r_a(s-a)$, where $s$ is the semiperimeter. Then, since $\triangle ADE\sim\triangle ABC$ and the $A-$excircle of $\triangle ADE$ is the incircle of $\triangle ABC$, we find that $\frac{DE}{BC}=\frac{r}{r_a}=\frac{s-a}{s}=\frac{b+c-a}{a+b+c}$. Thus, $DE=\frac{a(b+c-a)}{a+b+c}$. Finally, \begin{align*} (b+c-3a)^2&\ge0\\ 9a^2+b^2+c^2-6ab-6ca+2bc&\ge0\\ a^2+b^2+c^2+2ab+2ca+2bc&\ge8ab+8ca-8a^2\\ (a+b+c)^2&\ge8a(b+c-a)\\ a+b+c&\ge\frac{8a(b+c-a)}{a+b+c}\\ AB+BC+CA\ge8DE.\,\blacksquare \end{align*}