0<X<1,0<X<1怎样证√X²+Y² + √X²+(1-Y)² + √(1-X)²+Y² +√(1-X)²+(1-Y)² ≥2√2最好用牛逼点的方法.然后就是用对称性怎么证?
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/21 18:14:06
![0<X<1,0<X<1怎样证√X²+Y² + √X²+(1-Y)² + √(1-X)²+Y² +√(1-X)²+(1-Y)² ≥2√2最好用牛逼点的方法.然后就是用对称性怎么证?](/uploads/image/z/751368-48-8.jpg?t=0%EF%BC%9CX%EF%BC%9C1%2C0%EF%BC%9CX%EF%BC%9C1%E6%80%8E%E6%A0%B7%E8%AF%81%E2%88%9AX%26%23178%3B%2BY%26%23178%3B+%2B+%E2%88%9AX%26%23178%3B%2B%EF%BC%881-Y%EF%BC%89%26%23178%3B+%2B+%E2%88%9A%EF%BC%881-X%EF%BC%89%26%23178%3B%2BY%26%23178%3B+%2B%E2%88%9A%EF%BC%881-X%EF%BC%89%26%23178%3B%2B%EF%BC%881-Y%EF%BC%89%26%23178%3B+%E2%89%A52%E2%88%9A2%E6%9C%80%E5%A5%BD%E7%94%A8%E7%89%9B%E9%80%BC%E7%82%B9%E7%9A%84%E6%96%B9%E6%B3%95.%E7%84%B6%E5%90%8E%E5%B0%B1%E6%98%AF%E7%94%A8%E5%AF%B9%E7%A7%B0%E6%80%A7%E6%80%8E%E4%B9%88%E8%AF%81%3F)
0<X<1,0<X<1怎样证√X²+Y² + √X²+(1-Y)² + √(1-X)²+Y² +√(1-X)²+(1-Y)² ≥2√2最好用牛逼点的方法.然后就是用对称性怎么证?
0<X<1,0<X<1怎样证√X²+Y² + √X²+(1-Y)² + √(1-X)²+Y² +√(1-X)²+(1-Y)² ≥2√2
最好用牛逼点的方法.
然后就是用对称性怎么证?
0<X<1,0<X<1怎样证√X²+Y² + √X²+(1-Y)² + √(1-X)²+Y² +√(1-X)²+(1-Y)² ≥2√2最好用牛逼点的方法.然后就是用对称性怎么证?
由于√(x^2+y^2) >=√2/2*(x+y),后面的3项也用这个.得,原式>=√2/2*(x+y+x+1-y+1-x+y+1-x+1-y)=√2/2*4=2√2.
另外告诉你一个很有用的.
2/(1/x+1/y)