假设a1-a2+a3-a4+a5=2t，我们有
a1+a3+a5=1.5+t, a2+a4=1.5-t
12=(a1^2+a3^2+a5^2)+(a2^2+a4^2)>=(a1+a3+a5)^2/3+(a2+a4)^2/2
=5/6 t^2 - 1/2 t + 15/8，从而有
(3-6Sqrt(34))/10<=t<=(3+6Sqrt(34))/10
(3-6Sqrt(34))/5<=2t<=(3+6Sqrt(34))/5
两边取到等号时都要求a1=a3=a5，a2=a4，而3a1=1.5+t, 2a2=1.5-t，可以把取等条件解出来。
a1+a3+a5=1.5+t, a2+a4=1.5-t
12=(a1^2+a3^2+a5^2)+(a2^2+a4^2)>=(a1+a3+a5)^2/3+(a2+a4)^2/2
=5/6 t^2 - 1/2 t + 15/8，从而有
(3-6Sqrt(34))/10<=t<=(3+6Sqrt(34))/10
(3-6Sqrt(34))/5<=2t<=(3+6Sqrt(34))/5
两边取到等号时都要求a1=a3=a5，a2=a4，而3a1=1.5+t, 2a2=1.5-t，可以把取等条件解出来。
