Cho $\left\{ \begin{array}{l} n \in N\\ n\geq 2\end{array} \right.$ và $a_1;a_2;...;a_n \in [0;1].$
Chứng minh rằng:
                $\sum_{cyclic}^{} \frac{a_1}{a_2+a_3+....+a_{n}+1}+(1-a_1)(1-a_2).......(1-a_n)\leq 1 $
cho n=3 thì cái tổng hoán vị kia >= 3/2 (bdt nesbit ) :)) –  tran85295 05-07-16 07:31 AM
Ko mất tính tổng quát, giả sử $a_1=\max \lbrace a_1,a_2,a_3,...a_n \rbrace$
Khi đó 
$$\frac{a_2}{a_1+a_3+a_4+...+a_{n}+1} \le \frac{a_2}{a_2+a_3+a_4+...+a_{n}+1 }$$
$$\frac{a_3}{a_1+a_2+a_4+...+a_{n}+1} \le \frac{a_3}{a_2+a_3+a_4+...+a_{n}+1 }$$
$$\frac{a_4}{a_1+a_2+a_3+a_5+...+a_{n}+1} \le \frac{a_4}{a_2+a_3+a_4+...+a_{n}+1 }$$
$$\begin{matrix} ............\\ ............ \end{matrix}$$
$$\frac{a_{n+1}}{a_1+a_2+a_3+...+a_n} \le \frac{a_4}{a_2+a_3+a_4+...+a_{n}+1 }$$
Do đó $\sum_{cyc}\frac{a_1}{a_2+a_3+a_4+...+a_{n}+1} \le \frac{a_1+a_2+a_3...+a_{n}}{a_2+a_3+a_4+...+a_{n}+1}$
Nên chỉ cần cm 
$f(a_1)=\frac{a_1+a_2+a_3+...+a_{n}}{a_2+a_3+a_4+...+a_{n}+1}+\prod_{i=1}^n(a_i-1)-1 \le 0$
Biểu thức $f(a_1)$ là hàm bậc nhất theo biến $a_1\in[0;1]$
$\Rightarrow f(a_1) \le \max \lbrace {f(0);f(1)} \rbrace$
Lại có $f(1)=0,f(0)=0$ (do $a_1=\max \lbrace a_1,a_2,a_3,...a_n \rbrace$, và $a_1=0$)
$\Rightarrow $ dpcm
Đẳng thức xảy ra khi các biến nhận giá trị 0,1

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