Question

设△ABC的内角A,B,C的对边长分别为a,b,c,已知a,b,c成等比数列

Answer 1

1)要证a^2+b^2+c^2>(a-b+c)^2

只需证a^2+b^2+c^2>a^2+b^2+c^2-(2ab+2bc-2ac)

只需证2ab+2bc-2ac>0,ab+bc-ac>0

因为b^2=ac 所以只需证b(a+c)-b^2>0

只需证b(a+c-b)>0

只需证a+c-b>0 （三角形两边之和大于第三边）显然成立

所以a^2+b^2+c^2>(a-b+c)^2成立

2) ccos (A-C) + cos B = 3/2

所以 cos(A-C) + cos(π - A - C) = 3/2

cos(A-C) + cos(π - A - C) = cos(A - C ) - cos(A + C)

= cosAcosC + sinAsinC - cosAcosC + sinAsinC = 2sinAsinC = 3/2

所以 sinAsinC = 3/4

正弦定理

a/sinA = b/sinB =c/sinC,而b^2 = ac,即 b/a = c/b

所以 sinB/sinA = sinC/sinB,所以 sinB ^2 = sinAsinc

所以sinB^2 = 3/4,sinB = 根号3/2,B=60°或120°

但是,如果 B=60°,则cosB = -1/2

带入 cos(A-C)+cosB=3/2,cos(A-C) = 2,矛盾,所以舍去这个值.