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关于不等式问题的解法很多,构造函数是一种方法,现举数列如下: 例1 已知x、y、z∈(0,1),求证:x(1-y)+y(1-z)+z(1-x)<1. 猛一看,不知从哪儿入手,可我们仔细观察后发现,原式可变形为x(1-y)+y(1-z)+z(1-x)-1<0.我们可以将该式变成形如y=bx+b的直线,设f(x)=(1-y-z)x+(y-1+z-zy).想一想,只要线段的两个端点都在x轴下方,就有整个线段都在x轴下方了.由f(0)=(y-1)(1-z)<0,及f(1)=1-y-z+y+z-1-yz=-yz<0,可得原式成立.
There are many solutions to the problem of inequality. The constructor is a method. The current sequence is as follows: Example 1 Known x, y, z∈(0,1), Prove that: x(1-y)+y(1-z) +z(1-x)<1. When you look at it, you don’t know where to begin, but after careful observation, you can see that the original model can be transformed into x(1-y)+y(1-z)+z(1-x ) -1<0. We can change this formula to a straight line of the form y=bx+b. Let f(x)=(1-yz)x+(y-1+z-zy). Think of it as long as Both ends of the line segment are below the x-axis, and the entire line segment is below the x-axis. From f(0)=(y-1)(1-z)<0, and f(1)=1-y -z+y+z-1-yz=-yz<0, available original form is established.