1.
先设a1<a2<...<an,ak≠0.
记σ(n,k)=C(n,k)e(n,k),其中e(n,k)就是你定义的ek.
用归纳法证明:
e(n,k-1)e(n,k+1)≤e(n,k)^2,其中2≤k≤n-1.
ⅰ.
n=3,
F(T)=(a1T+1)(a2T+1)(a3T+1)=
=σ(3,3)T^3+σ(3,2)T^2+σ(3,1)T+1
由于F(T)有3个不同的实根,所以根据导数的中值定理得,
F'(T)有2个不同的实根,
==>
[2σ(3,2)]^2-4[3σ(3,3)][σ(3,1)]
==>
e(3,1)e(3,3)≤e(3,2)^2
ⅱ.
设命题对n时成立.
则G(T,n)=(T+a1)(T+a2)..(T+an)=
=T^n+C(n,1)e(n,1)T^(n-1)+...+C(n,n)e(n,n)
有e(n,k-1)e(n,k+1)≤e(n,k)^2.
设G(T,n+1)=(T+a1)(T+a2)..(T+an)(T+a(n+1))=
=T^(n+1)+C(n+1,1)e(n+1,1)T^n+...+C(n+1,n+1)e(n+1,n+1)
由于G(T,n+1)有n+1个不同的实根,所以根据导数的中值定理得,
G'(T,n+1)有n个不同的实根.
==>
G'(T,n+1)=
=(n+1)T^n+nC(n+1,1)e(n+1,1)T^(n-1)+...+C(n+1,n)e(n+1,n)
=(n+1)[T^n+C(n,1)e(n+1,1)T^(n-1)+...+C(n,n)e(n+1,n)]
根据假设得:
e(n,k-1)e(n,k+1)≤e(n,k)^2,1≤k≤n-1.
设
F(T,n+1)=(a1T+1)(a2T+1)..(a(n+1)T+1)=
=C(n+1,n+1)e(n+1,n+1)T^(n+1)+C(n+1,n)e(n+1,n)T^(n)+
+C(n+1,n-1)e(n+1,n-1)T^(n-1)+....C(n+1,1)e(n+1,1)T+1
==>
F^((n-1))(T,n+1)=[(n+1)!/2]C(n+1,n+1)e(n+1,n+1)T^2+
+n!C(n+1,n)e(n+1,n)T+(n-1)!C(n+1,n-1)e(n+1,n-1)
=[(n+1)!/2][e(n+1,n+1)T^2+2e(n+1,n)T+e(n+1,n-1)]
有2个不同的实根.
==>
[2e(n+1,n)]^2-4e(n+1,n+1)e(n+1,n-1)>0
==>
命题对n+1时成立.
所以当a1<a2<...<an,ak≠0,命题成立.
2.
若a1≤a2≤...≤an,
则当m>M时,a1+1/m<a2+2/m<...<an+n/m,ak+k/m≠0
根据1.不等式对于a1+1/m,a2+2/m,...,an+n/m成立.
然后取m→∞,使不等式对于a1,a2,...,an成立.
