Cho \(n \in {\rm N},\,\,n \ge 2\). Chứng minh rằng \(C_n^0C_n^1C_n^2...C_n^n \le {\left( {\frac{{{2^n} - 2}}{{n - 1}}} \right)^{n - 1}}\).

Phương pháp:

Áp dụng: \[C_n^0 + C_n^1 + C_n^2 + ... + C_n^n = {2^n}\] và BĐT Cô si

Cách giải:

Ta có: \[C_n^0 + C_n^1 + C_n^2 + ... + C_n^n = {2^n} \Leftrightarrow C_n^1 + C_n^2 + ... + C_n^{n - 1} = {2^n} - 2\]

Áp dụng BĐT Cô si cho \[n - 1\] số dương \[C_n^1,\,C_n^2,\,...,\,C_n^{n - 1}\] ta có:

\[C_n^1 + C_n^2 + ... + C_n^{n - 1} \ge \left( {n - 1} \right)\sqrt[{n - 1}]{{C_n^1C_n^2\,...C_n^{n - 1}}},\,\forall n \in \mathbb{N},\,n \ge 2\]

\[ \Rightarrow {2^n} - 2 \ge \left( {n - 1} \right)\sqrt[{n - 1}]{{C_n^1C_n^2\,...C_n^{n - 1}}} \Leftrightarrow C_n^1C_n^2\,...C_n^{n - 1} \le {\left( {\frac{{{2^n} - 2}}{{n - 1}}} \right)^{n - 1}} \Leftrightarrow C_n^0C_n^1C_n^2\,...C_n^{n - 1}C_n^n \le {\left( {\frac{{{2^n} - 2}}{{n - 1}}} \right)^{n - 1}}\]

(do \[C_n^0 = C_n^n = 1\])

Vậy, \[C_n^0C_n^1C_n^2\,...C_n^n \le {\left( {\frac{{{2^n} - 2}}{{n - 1}}} \right)^{n - 1}},\,\forall n \in \mathbb{N},\,n \ge 2\].