sum(de i=1 a n-1)(bi+1²)/ai*sum(de i=1 a n-1)ai >=[sum(de i=1 a n-1) bi+1]² ( C-S)
on a
sum(de i=1 a n-1)ai =1
on deduit que
sum(de i=1 a n-1)(bi+1²)/ai>=[sum(de i=1 a n-1) bi+1]²
on a sum(de i=1 a n-1)bi+1=sum(de i=2 a n)bi
et
sum(de i=1 a n-1)(bi+1²)+b1²>=[sum(de i=2 a n)bi]²+b1²
>=2b1(sum i=1 a n)bi(a²+b²>=2ab)
CQFD
a+