已知三角形ABC的三邊長a,b,c滿足b+c≤2a,c+a≤2b,則(a+b)²/ab的取值范圍是?

最小值是a=b=c時(shí)候,得到最小值4

最大比較麻煩：

利用內(nèi)切圓切線長代換

令

a=x+y

b=y+z

c=z+x

x,y,z>0

設(shè)x+y+z=1

b+c≤2a

2z≤x+y

c+a≤2b

2x≤y+z

M=(a+b)²/ab

=(x+2y+z)²/(y(x+y+z)+xz)

=(1+y)²/(y+xz)

當(dāng)y固定時(shí),

x+z=1-y也固定

2z≤x+y

2x≤z+y

所以當(dāng)2/3>=y>1/3時(shí)候

xz>=(1/3)(-y+2/3)

M=(1+y)²/(y+xz)

<=(1+y)²/(y+(1/3)(-y+2/3))

=(3/2)(1+y)²/(y+1/3))

令y+1/3=t

1>=t>2/3

(3/2)(1+y)²/(y+1/3))

=(3/2)(t+2/3)²/t

=2+(3/2)[t+(4/9)(1/t)]

<=2+13/6

=25/6

當(dāng)1>=y>2/3時(shí)候

xz>=0

M=(1+y)²/(y+xz)

<=(1+y)²/y=2+y+1/y<=2+2/3+3/2=25/6

但是如果y=2/3則要求x,z之一為0,所以25/6是上限,但是取不到.

所以綜上,所求的范圍是：

[4,25/6)