证明 根据Erdos-Mordell不等式:PA+PB+PC≥2(PD+PE+PF)记 M=PA+PB+PC,N=PD+PE+PF所以 3(M-N)≥M+N<==>3(M-N)^2≥M^2-N^2. (1)根据简单三角形不等式:(BP+CP)^2≥BC^2+4PD^2=a^2+4PD^2;(CP+AP)^2≥CA^2+4PE^2=b^2+4 展开
证明 根据Erdos-Mordell不等式:PA+PB+PC≥2(PD+PE+PF)记 M=PA+PB+PC,N=PD+PE+PF所以 3(M-N)≥M+N<==>3(M-N)^2≥M^2-N^2. (1)根据简单三角形不等式:(BP+CP)^2≥BC^2+4PD^2=a^2+4PD^2;(CP+AP)^2≥CA^2+4PE^2=b^2+4PE^2;(A{+BP)^2≥AB^2+4PF^2=c^2+4PF^2.由柯西不等式得:4(PA+PB+PC)^2=[(PB+PC)+(PC+PA)+(PA+PB)]^2=(PB+PC)^2+(PC+PA)^2+(PA+PB)^2+2(PC+PA)*(PA+PB)+2(PA+PB)*(PB+PC)+2(PB+PC)*(PC+PA)≥a^2+4PD^2+b^2+4PE^2+b^2+4PF^2+2(bc+4PE*PF+ca+4PF*PD+ab+4PD*PE)=(a+b+c)^2+4(PD+PE+PF)^2即得:M^2≥s^2+N^2<==>M^2-N^2≥s^2 (2)由(1),(2)得:3(T-t)^2≥s^2 因此 PA+PB+PC-(PD+PE+PF)≥s/√3. 收起
