论文部分内容阅读
若n∈N,n>1,则(1+x)~n≥1+nx. 其中等号当且仅当x=0时成立. 这就是著名的贝努利不等式.高中《代数》下册第123页用数学归纳法给出了它的证明,但未介绍它的应用.本文兹举几例,供教学时参考. 例1若x_i>-1,(i+1,2,…,n),n∈N,且x_1+x_2+…+x_n=0,求证 (1+x_1)~n+(1+x_2)~n+…+(1+x_n)~n≥n. 证明:当n=1时,等号显然成立. 当n>1时,由贝努利不等式知(1+x_1)~n+(1+x_2)~n+…+(1+x_n)~n≥(1+nx_1)+(1+nx_2)+…+(1+nx_n)=n+n(x_1+x_2+…+x_n)=n.
If n∈N,n>1, then (1+x)~n≥1+nx. The equal sign holds if and only if x=0. This is the famous Bernoulli inequality. Page 123 gives its proof using mathematical induction, but does not introduce its application. Here are some examples for reference when teaching. Example 1 If x_i > -1, (i+1,2,...,n ), n∈N, and x_1+x_2+...+x_n=0, verify (1+x_1)~n+(1+x_2)~n+...+(1+x_n)~n≥n. Proof: When n=1 The equal sign is obviously established. When n>1, we know by Bernoulli’s inequality that (1+x_1)~n+(1+x_2)~n+...+(1+x_n)~n≥(1+nx_1)+(1 +nx_2)+...+(1+nx_n)=n+n(x_1+x_2+...+x_n)=n.