Question

设a>0,b>0,c>0 ,且满足a^2=b(b+c),b^2=c(c+a) 求证:1/a+1/b=1/c

Answer 1

结论可化为c
=
(ab)/(a+b)
a^2
=
b^2+bc
=
c^2+ac+bc
=>(a+c)(a-c)
=
c(a+b)
(1)
由b^2
=
c(a+c)
=>
a+c
=
b^2/c
代入（1）得
b^2(a-c)
=
c^2(a+b)
整理为c的二次式得(a+b)c^2+b^2*c-a*b^2
=
0
按c分解因式得[(a+b)c-ab](c+b)
=
0
因为c>0,b>0,所以b+c>0
所以(a+b)c-ab=0
即得1/a+1/b=1/c