Forgot password
 Register account
View 1803|Reply 8

[函数] 一个函数问题2016_5_11

[Copy link]

412

Threads

1432

Posts

3

Reputation

Show all posts

realnumber posted 2016-5-11 10:44 |Read mode
辽宁庄河高中朱--
$f(x)=\frac{e^x}{x},f(m)=f(n),求\frac{1}{n}+\frac{1}{m}$的取值范围.


先放个可能相关的连接
forum.php?mod=viewthread&tid=2497&highlight=

673

Threads

110K

Posts

218

Reputation

Show all posts

kuing posted 2016-5-11 11:29
打漏了 $m\ne n$

412

Threads

1432

Posts

3

Reputation

Show all posts

original poster realnumber posted 2016-5-11 11:52
回复 2# kuing


    en ,应该加上,或换为m,n>0

412

Threads

1432

Posts

3

Reputation

Show all posts

original poster realnumber posted 2016-5-11 12:05
似乎这样可以
不妨设$0<m\le1,n\ge 1$,
利用$e^x\ge 1+x,e^x \ge 1+x+x^2/2,x\ge 0$
$e^{n-m}=\frac{n}{m}\ge 1+n-m+\frac{(n-m)^2}{2},得到\frac{1}{m}\ge 1+\frac{n-m}{2}$
即$n\le \frac{2}{m}+m-2$
可得$\frac{1}{m}+\frac{1}{n}\ge \frac{1}{m}+\frac{m}{m^2-2m+2}\ge 2$.(最后是减函数,求导)

412

Threads

1432

Posts

3

Reputation

Show all posts

original poster realnumber posted 2016-5-11 12:37
张小明老师的办法
未命名8678678.JPG

764

Threads

4672

Posts

27

Reputation

Show all posts

isee posted 2016-5-11 14:05
又是楼主智囊团解决了。。。。。。

412

Threads

1432

Posts

3

Reputation

Show all posts

original poster realnumber posted 2016-5-11 15:12
回复 6# isee


    群里在发,有人解答,就发上来了。
什么团,没有吧。

7

Threads

578

Posts

9

Reputation

Show all posts

游客 posted 2016-5-11 16:30
作图其实很显然。

54

Threads

160

Posts

0

Reputation

Show all posts

血狼王 posted 2016-5-11 16:46
事实上有更强结论:
$$3(m+n)^3\geq 8(m^2+mn+n^2).$$
大家不妨一试。

Quick Reply

Advanced Mode
B Color Image Link Quote Code Smilies
You have to log in before you can reply Login | Register account

$\LaTeX$ formula tutorial

Mobile version

2025-7-15 14:01 GMT+8

Powered by Discuz!

Processed in 0.021765 seconds, 25 queries