若n为自然数,且1/a+1/b+1/c=1/a+b+c,求证:1/(a)2n+1+1/(b)2n+1+1/(c)2n+1

∵1/a+1/b+1/c=1/(a+b+c)
(bc+ac+ab)/(abc)=1/(a+b+c)
(bc+ac+ab)(a+b+c)=abc
abc+b^2c+bc^2+a^2c+abc+ac^2+a^2b+ab^2+abc=abc
b^2c+bc^2+a^2c+ac^2+a^2b+ab^2+2abc=0
(b^2c+a^2b+ab^2+abc)+(bc^2+a^2c+ac^2+abc)=0
b(bc+a^2+ab+ac)+c(bc+a^2+ac+ab)=0
(b+c)(bc+a^2+ab+ac)=0
(b+c)[(bc+ab)+(a^2+ac)]=0
(b+c)[b(a+c)+a(a+c)]=0
(b+c)(b+a)(a+c)=0
∴a+b=0或b+c=0或c+a=0,
即a=-b或b=-c或c=-a.
∵2n+1是奇数,∴当a=-b时
1/(a)2n+1,1/(b)2n+1互为相反数
1/(a)2n+1+1/(b)2n+1+1/(c)2n+1=1/(a)2n+1+(b)2n+1+(c)2n+1
左边
=1/(a)2n+1+1/(b)2n+1+1/(c)2n+1
=0+1/(c)2n+1
=1/(c)2n+1
右边
=1/(a)2n+1+(b)2n+1+(c)2n+1
=1/0+(c)2n+1
=1/(c)2n+1
∴左边=右边
∴1/(a)2n+1+1/(b)2n+1+1/(c)2n+1=1/(a)2n+1+(b)2n+1+(c)2n+1
同理当b=-c、当c=-a时,也成立
1/(a)2n+1+1/(b)2n+1+1/(c)2n+1=1/(a)2n+1+(b)2n+1+(c)2n+1