已知$\Delta{ABC},AB=c,BC=a,CA=b,P$為平面內任意一點.證明:
$(PA+PB+PC)^2\ge\sqrt{3}(a\cdot PA+b\cdot PB+c\cdot PC)$
由1971年加拿大M.S.Klamkin教授給出的不等式:
$(\lambda_1+\lambda_2+\lambda_3)(\lambda_1PA^2+\lambda_2PB^2+\lambda_3PC^2)\ge\lambda_2\lambda_3a^2+\lambda_3\lambda_1b^2+\lambda_1\lambda_2c^2$
令$\lambda_1=\dfrac{1}{PA},\lambda_2=\dfrac{1}{PB},\lambda_3=\dfrac{1}{PC}$
得$(PB\cdot PC+PC\cdot PA+PA\cdot PB)(PA+PB+PC)\ge a^2PA+b^2PB+c^2PC$
又$(PA+PB+PC)^2\ge3(PB\cdot PC+PC\cdot PA+PA\cdot PB)$
故$(PA+PB+PC)^4\ge3(PB\cdot PC+PC\cdot PA+PA\cdot PB)(PA+PB+PC)^2$
$\ge3(a^2PA+b^2PB+c^2PC)(PA+PB+PC)$
$\ge3(aPA+bPB+cPC)^2$
即得$(PA+PB+PC)^2\ge\sqrt{3}(a\cdot PA+b\cdot PB+c\cdot PC)$