Forgot password
 Register account
View 315|Reply 2

[不等式] 最小值问题

[Copy link]

209

Threads

949

Posts

2

Reputation

Show all posts

敬畏数学 posted 2024-10-4 10:30 |Read mode
正数x,y满足$ \sqrt{9x^2-1}+ \sqrt{9y^2-1}=9xy, $则$ 4x^2+y^2 $最小值——————。

673

Threads

110K

Posts

218

Reputation

Show all posts

kuing posted 2024-10-4 13:36
令 `\sqrt{9x^2-1}=a`, `\sqrt{9y^2-1}=b`,则 `x=\frac13\sqrt{1+a^2}`, `y=\frac13\sqrt{1+b^2}`,条件变为
\begin{gather*}
a+b=\sqrt{(1+a^2)(1+b^2)},\\
(a+b)^2=(1+a^2)(1+b^2),\\
(ab-1)^2=0,\\
ab=1,
\end{gather*}
所以
\[4x^2+y^2=\frac 49(1+a^2)+\frac 19(1+b^2)=\frac 19(5+4a^2+b^2)\geqslant \frac 19(5+4ab)=1.\]

209

Threads

949

Posts

2

Reputation

Show all posts

original poster 敬畏数学 posted 2024-10-4 14:08
kuing 发表于 2024-10-4 13:36
令 `\sqrt{9x^2-1}=a`, `\sqrt{9y^2-1}=b`,则 `x=\frac13\sqrt{1+a^2}`, `y=\frac13\sqrt{1+b^2}`,条件变 ...
谢谢大师提醒。看到倒数第三行好像直接可以柯西(或者向量)!

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

2025-7-20 13:41 GMT+8

Powered by Discuz!

Processed in 0.018182 seconds, 45 queries