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利用配方法容易证明下面的代数恒等式:a~2+b~2+c~2-ab-bc-ca=1/2[(a-b)~2+(b-c)~2+(c-a)~2](*)上式左右两边关于 a、b、c 轮换对称,并且右边是三个非负数之和的一半,同时隐含着(a-b)+(b-c)+(c-a)=0这一条件.下面举例说明它在解一类竞赛题中的应用.
It is easy to prove the following algebraic identities using the matching method: a~2+b~2+c~2-ab-bc-ca=1/2[(ab)~2+(bc)~2+(ca)~2] (*) The left and right sides of the above equation are rotationally symmetrical about a, b, and c, and the right side is half of the sum of the three non-negative numbers, and the condition (ab)+(bc)+(ca)=0 is implied. An illustration of its application in solving a class of competition problems.