已知$Delta{ABC},AB=c,BC=a,CA=b,P$为平面内任意一点.证明:
$(PA+PB+PC)^2gesqrt{3}(acdot PA+bcdot PB+ccdot PC)$
由1971年加拿大M.S.Klamkin教授给出的不等式:
$(lambda_1+lambda_2+lambda_3)(lambda_1PA^2+lambda_2PB^2+lambda_3PC^2)gelambda_2lambda_3a^2+lambda_3lambda_1b^2+lambda_1lambda_2c^2$
令$lambda_1=dfrac{1}{PA},lambda_2=dfrac{1}{PB},lambda_3=dfrac{1}{PC}$
得$(PBcdot PC+PCcdot PA+PAcdot PB)(PA+PB+PC)ge a^2PA+b^2PB+c^2PC$
又$(PA+PB+PC)^2ge3(PBcdot PC+PCcdot PA+PAcdot PB)$
故$(PA+PB+PC)^4ge3(PBcdot PC+PCcdot PA+PAcdot PB)(PA+PB+PC)^2$
$ge3(a^2PA+b^2PB+c^2PC)(PA+PB+PC)$
$ge3(aPA+bPB+cPC)^2$
即得$(PA+PB+PC)^2gesqrt{3}(acdot PA+bcdot PB+ccdot PC)$