等式的证明
1个回答

∵1/a+1/b+1/c = 1/(a+b+c),

∴(ab+bc+ca)(a+b+c) = abc,

∴(a+b)(b+c)(c+a) = a²b+b²c+c²a+ab²+bc²+ca²+2abc = 0,

∴a+b,b+c,c+a中至少有一个是0.

由对称性,不妨设b+c = 0,即b = -c.

则对任意奇数n,有b^n = (-c)^n = -c^n.

因此1/a^n+1/b^n+1/c^n = 1/a^n+1/b^n-1/b^n = 1/a^n = 1/(a+b+c)^n.