本文主要是介绍＜Senior High School Math＞: inequality question，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
( 1 ) . o m i t (1). omit (1).omit
( 2 ) . ( a 2 − b 2 ) ( x 2 a 2 − y 2 b 2 ) = ( x 2 + y 2 ) − ( a 2 y 2 b 2 + b 2 x 2 a 2 ) ≤ x 2 + y 2 − 2 x y = ( x − y ) 2 (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.
( 3 ) . (3). (3). s e t : 3 m − 5 = x , m − 2 = y set: \sqrt{3m-5}=x,\sqrt{m-2}=y set:3m−5=x,m−2=y
t h e n : x 2 − 3 y 2 = 1 then: x^2-3y^2=1 then:x2−3y2=1
h a v e : x 2 1 − y 2 1 3 = 1 , have:\frac{x^2}{1}-\frac{y^2}{\frac{1}{3}}=1, have:1x2−31y2=1, according to the question(2), we get:
( x − y ) 2 ≥ ( 1 − 1 3 ) ( x 2 1 − y 2 1 3 ) = 2 3 (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.
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