＜Senior High School Math＞: inequality question
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t
(1). omit
(1).omit
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=
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≤
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(2). (a^2-b^2)(\frac{x^2}{a^2} - \frac{y^2}{b^2})=(x^2+y^2)-(\frac{a^2y^2}{b^2}+\frac{b^2x^2}{a^2}) \le x^2+y^2-2xy=(x-y)^2
(2).(a2−b2)(a2x2−b2y2)=(x2+y2)−(b2a2y2+a2b2x2)≤x2+y2−2xy=(x−y)2
- when a y b = b x a \frac{ay}{b}=\frac{bx}{a} bay=abx, the equation is satisfied.
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(3).
(3).
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:
3
m
−
5
=
x
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m
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2
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y
set: \sqrt{3m-5}=x,\sqrt{m-2}=y
set:3m−5=x,m−2=y
t
h
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n
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x
2
−
3
y
2
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1
then: x^2-3y^2=1
then:x2−3y2=1
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v
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:
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1
−
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2
1
3
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1
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have:\frac{x^2}{1}-\frac{y^2}{\frac{1}{3}}=1,
have:1x2−31y2=1, according to the question(2), we get:
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(x-y)^2 \ge (1-\frac{1}{3})(\frac{x^2}{1}-\frac{y^2}{\frac{1}{3}})=\frac{2}{3}
(x−y)2≥(1−31)(1x2−31y2)=32
- Pay attention to equal conditions!!!
- we can click here to see some more examples about the Cauchy-inequality.
原文地址：https://blog.csdn.net/weixin_46620878/article/details/136724032
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