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勾股定理是大家所熟知的:在△ABC中,∠C=90°,则c~2=a~2+b~2。这是直角三角形中的边的平方关系,在边的一次关系中还有ca~3+b~3这个不等式怎样证明呢? 注意到这是在直角三角形中的关系,应有c>a,c>b和c~3=a~3+b~2,所以 c~3=ca~2+cb~2>a·a~2+b·b~2=a~3+b~3 我们看到证明是比较简单的,如果你是独立完成
The Pythagorean theorem is well-known: In △ABC, ∠C=90°, then c~2=a~2+b~2. This is the square relationship of the sides in a right-angled triangle. In the first-order relationship of edges, there is ca~3+b~3 prove? Note that this is a relationship in a right triangle. There should be c>a,c>b and c~3=a~3+b ~2, so c~3=ca~2+cb~2>a·a~2+b·b~2=a~3+b~3 We see that the proof is relatively simple if you are done independently