我按楼主自己的思路做的,证明AB+AE < BE+AD
设BD=1,AD=k
则DE=k^2,AB=√(k^2+1),AE=k√(k^2+1)
∴BE+AD = 1+k+k^2
(BE+AD)^2 = (1+k+k^2)^2 = 1+2k+3k^2+2k^3+k^4
AB+AE = (k+1)√(k^2+1)
(AB+AE)^2 = (k+1)^2 * (k^2+1) = 1+2k+2k^2+2k^3+k^4
我按楼主自己的思路做的,证明AB+AE < BE+AD
设BD=1,AD=k
则DE=k^2,AB=√(k^2+1),AE=k√(k^2+1)
∴BE+AD = 1+k+k^2
(BE+AD)^2 = (1+k+k^2)^2 = 1+2k+3k^2+2k^3+k^4
AB+AE = (k+1)√(k^2+1)
(AB+AE)^2 = (k+1)^2 * (k^2+1) = 1+2k+2k^2+2k^3+k^4