a^2+b^2+c^2+d^2-ab-bc-cd-da
=1/2*(2a^2+2b^2+2c^2+2d^2-2ab-2bc-2cd-2da)
=1/2*(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2cd+d^2+a^2-2da+d^2)
=1/2*[(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2]
因为(a-b)^2>=0,(b-c)^2>=0,(c-d)^2>=0,(d-a)^2>=0,
所以1/2*[(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2]>=0
即a^2+b^2+c^2+d^2-ab-bc-cd-da>=0