2.cho I là tâm đường tròn nội tiếp tam giác ABC có diên tích S và nửa chu vi p chứng minh IA+IB+IC $\geq \frac{6S}{p}$
Lời giải. Gọi độ dài ba cạnh tam giác $ABC$ là $a,b,c$. Ta có $p= \frac{a+b+c}{2}$. Gọi $r$ là bán kính đường tròn nội tiếp tam giác $ABC$.
Theo định lý Pythagores thì $$AI= \sqrt{r^2+ \left( \frac{b+c-a}{2} \right)^2}, IB= \sqrt{r^2+ \left( \frac{a+c-b}{2} \right)^2}, \\ IC= \sqrt{r^2+ \left( \frac{a+b-c}{2} \right)^2}.$$
Áp dụng BĐT Cauchy-Schwarz ta có $$\begin{array}{l} (1+3) \left[ r^2+ \left( \frac{a+b-c}{2} \right)^2 \right] \ge \left( r+ \frac{(a+b-c) \cdot \sqrt 3}{2} \right)^2 \\ \Rightarrow 2CI \ge r+ \frac{(a+b-c) \cdot \sqrt 3}{2}. \end{array}$$
Chứng minh tương tự ta được $2(AI+BI+CI) \ge 3r+ \frac{ \sqrt 3(a+b+c)}{2}$.
Mặt khác $r \cdot p=S$ nên $r= \frac{S}{p}$ suy ra $\frac{6S}{p}=6r$. Ta cần chứng minh
$$\begin{aligned} \sqrt 3(a+b+c) \ge 18r & \Leftrightarrow (a+c+b)^2 \ge 12 \sqrt 3S \\ & \Leftrightarrow (a+b+c)^2 \ge 3 \sqrt{3(a+b+c)(a+b-c)(b+c-a)(c+a-b)} \\ & \Leftrightarrow (a+b+c)^3 \ge 27(a+b-c)(b+c-a)(c+a-b) \end{aligned}$$
Dễ chứng minh được rằng $(a+b-c)(b+c-a)(c+a-b) \le abc \le \frac{(a+b+c)^3}{27}$ nên $\frac{ \sqrt 3(a+b+c)}{2} \ge 9r$ suy ra $IA+IB+IC \ge 6r= \frac{6S}{p}$.
Dấu đẳng thức xảy ra khi và chỉ khi $\triangle ABC$ đều. $\blacksquare$
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).