已知abc均为非零实数,且1/a,1b 1/c成等差数列求证 (b+c)/a (c+a)/b (a+b)/c 为等差
已知abc均为非零实数,且1/a,1b 1/c成等差数列求证 (b+c)/a (c+a)/b (a+b)/c 为等差
已知abc均为非零实数,且1/a,1b 1/c成等差数列
求证 (b+c)/a (c+a)/b (a+b)/c 为等差
已知abc均为非零实数,且1/a,1b 1/c成等差数列求证 (b+c)/a (c+a)/b (a+b)/c 为等差
1/a,1/b 1/c成等差数列
2/b=1/a+1/c=(a+c)/(ac)
b(a+c)=2ac
(b+c)/a + (a+b)/c
=[(b+c)c+(a+b)a]/(ac)
=[a^2+c^2+b(a+c)]/(ac)
=(a^2+c^2+2ac)/(ac)
=(a+c)*(a+c)/(ac)
=(a+c)*2/b
=2(a+c)/b
所以(b+c)/a (c+a)/b (a+b)/c 为等差
1/a 1/b 1/c为等差数列,所以1/a+1/c=2/b
所以(a+b+c)/a+(a+b+c)/c=2(a+b+c)/b
所以(b+c)/a+1+(a+b)/c+1=2(a+c)/b+2
所以(b+c)/a+(a+b)/c=2(a+c)/b
所以后三者也是等差数列
1/a,1/b 1/c成等差数列 <==> 2/b = 1/a + 1/c
2/b = 1/a + 1/c ==> (b+c)/a + (a+b)/c = c/a + a/c + b(1/a + 1/c) = 2 + c/a + a/c
= (a+c)(1/a + 1/c) = (a+c)(2/b) = 2(c+a)/b
==> (b+c)/a, (c+a)/b, (a+b)/c 为等差数列