Forgot password
 Register account
View 183|Reply 3

[不等式] 怎么用不等式法求证等式

[Copy link]

16

Threads

18

Posts

0

Reputation

Show all posts

Canhuang posted 2023-3-11 21:00 from mobile |Read mode
$\frac{\sin^4A}{\sin^2B}+\frac{\cos^4A}{\cos^2B}=1$,证明$\sin^2A=\sin^2B$.

673

Threads

110K

Posts

218

Reputation

Show all posts

kuing posted 2023-3-11 21:33
很简单啊,由均值
\begin{align*}
\frac{\sin^4A}{\sin^2B}+\sin^2B&\geqslant2\sin^2A,\\
\frac{\cos^4A}{\cos^2B}+\cos^2B&\geqslant2\cos^2A,
\end{align*}
相加结合条件变成 `1+1\geqslant2`,那只能取等,即得 `\sin^2A=\sin^2B`。

764

Threads

4672

Posts

27

Reputation

Show all posts

isee posted 2023-3-11 21:46
由Cauchy 不等式分式形式\[1=\frac{\big(\sin^2A\big)^2}{\sin^2B}+\frac{\big(\cos^2A\big)^2}{\cos^2B}\geqslant \frac{\big(\sin^2A+\cos^2A\big)^2}{\sin^2B+\cos^2B}=1,\]
于是仅等号成立\[\frac{\sin^2A}{\sin^2B}=\frac{\cos^2A}{\cos^2B}=\frac{\sin^2A+\cos^2B}{\sin^2A+\cos^2B}=1.\]
isee=freeMaths@知乎

673

Threads

110K

Posts

218

Reputation

Show all posts

kuing posted 2023-3-11 22:31
isee 发表于 2023-3-11 21:46
由Cauchy 不等式分式形式\[1=\frac{\big(\sin^2A\big)^2}{\sin^2B}+\frac{\big(\cos^2A\big)^2}{\cos^2B}\g ...
我第一反应也是这个,后来觉得还是用均值“初级”一点更容易被接受😏

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 14:38 GMT+8

Powered by Discuz!

Processed in 0.013119 seconds, 22 queries