[tatari/nightmare]

ข้อความเดิมเขียนโดยคุณ kanakon

Let $n\geq 3$ be an integer. Let $t_1,t_2,t_3,...t_n$ be posiive real integer such that
$$(n^2+1)>(t_1+t_2+...+t_n)(\frac{1}{t_1}+\frac{1}{t_2}+...+\frac{1}{t_n} )$$
Show that $t_i,t_j,t_k$ are side lenghts of a triangle for all $i,j,k$ with $1\leq i<j<k\leq n$ [IMO 2004]

$(t_1+t_2+...+t_n)(\frac{1}{t_1}+\frac{1}{t_2}+...+\frac{1}{t_n})=n+\sum_{i<j}^{}(\frac{t_i}{t_j}+\frac{t_j}{t_i})
=n+t_1(\frac{1}{t_2}+\frac{1}{t_3})+\frac{1}{t_1}(t_2+t_3)+\sum_{i<j,(i,j)\not\in{(1,2),(1,3)}}^{}(\frac{t_i}{t_j}+\frac{t_j}{t_ i})$
จาก $$\frac{1}{t_2}+\frac{1}{t_3}\geq\frac{2}{\sqrt{t_{2}t_{3}}}$$
$$t_2+t_3\geq 2\sqrt{t_{2}t_{3}}$$ ( AM-GM)
และ $\frac{t_i}{t_j}+\frac{t_j}{t_i}\geq2$ ให้ $a=\frac{t_1}{\sqrt{t_{2}t_{3}}}$ ได้ $a>0$ จากโจทย์ได้
$n^2+1>(t_1+t_2+...+t_n)(\frac{1}{t_1}+\frac{1}{t_2}+...+\frac{1}{t_n})\geq n+\frac{2t_1}{\sqrt{t_{2}t_{3}}}+\frac{2\sqrt{t_{2}t_{3}}}{t_1}+2[\binom{n}{2}-2]=2a+\frac{2}{a}+n^2-4$
$\therefore 2a+\frac{2}{a}-5< 0$ นั่นคือ $2a^2-5a+2<0$ จะได้ $\frac{1}{2}<a<2$
นั่นคือ $\frac{t_1}{\sqrt{t_{2}t_{3}}}=a<2$
$\therefore t_1<2\sqrt{t_{2}t_{3}}$ แต่จาก $2\sqrt{t_{2}t_{3}}\leq t_2+t_3$
ฉะนั้น $t_1<t_2+t_3$