设P是△ABC平面上任一点,BC=a,CA=b,AB=c.求
M=(PB+PC)/(b+c)+(PC+PA)/(c+a)+(PA+PB)/(a+b)的最小值.

设P是△ABC平面上任一点,设BC=a,CA=b,AB=c.求
M=(PB+PC)/(b+c)+(PC+PA)/(c+a)+(PA+PB)/(a+b)的最小值.

解 设P点到边BC,CA,AB的距离为x,y,z.△,s,R,r分别表示△ABC的面积,半周长,外接与内切圆半径.
易证
(PB+PC)^2≥a^2+4x^2; (1-1)
(PC+PA)^2≥b^2+4y^2; (1-2)
(PA+PB)^2≥c^2+4z^2. (1-3)
和
bcx+cay+abz>2(4R+r)r (2)
显然可证:
bc>(4R+r)r, <==>(b+c-a)^2>0
ca>(4R+r)r, <==>(c+a-b)^2>0
ab>(4R+r)r. <==>(a+b-c)^2>0
所以(2)成立.
注意到面积恒等式:
ax+by+cz=2△

根据柯西不等式:
[∑(PB+PC)/(b+c)]^2≥[∑√(a^2+4x^2)/(b+c)]^2
≥[∑a/(b+c)]^2+4[∑x/(b+c)]^2
>{[∑a^3+∑a*∑bc]^2+4[4rs^2+2(4R+r)r^2]^2}/∏(b+c)^2

因为4R+r≥√3*s.所以只需证:
9[∑a^3+∑a*∑bc]^2+144r^2*[2s^2+(4R+r)r]^2>25∏(b+c)^2

∵[2s^2+(4R+r)r]^2≥4s^4+(16R+7r)rs^2
9[∑a^3+∑a*∑bc]^2+144r^2*s^2*[4s^2+(16R+7r)r]>25∏(b+c)^2

<===>
36(s^2-Rr-r^2)^2+36[4s^2+(16R+7r)r]r^2>25(s^2+2Rr+r^2)
^2

<==>
11s^4-172Rrs^2+22r^2*s^2-r^2(64R^2-548Rr-263r^)>0

<===>
(11s^2+4Rr-33r^2)*(s^2-16Rr+5r^2)+428r^4>0