x≤k(x)≤1/2(x^2+1)
x=1
1≤k(1)≤1
k(1)=1
y=ax^3/3+1/2bx^2+cx
x≤y'=ax^2+bx+c≤1/2(x^2+1)
k(-1)=0
a-b+c=0
k(1)=1
a+b+c=1
x≤k(x)
k(1)'=1
k(x)'=2ax+b
2a+b=1
a=1/4
b=1/2
c=1/4
k(x)=(1/4)(x+1)^2
1/k(1)+1/k(2)+……+1/k(n)>2n/(n+2)
1/4+1/9+1/16+……+1/(n+1)^2>n/(2n+4)(除以了4)
1/4+1/12+1/20+1/30+……+1/(n+1)(n+2)> n/(2n+4)
1/4+1/3-1/4+1/4-1/5……+1/(n+1)-1/(n+2)>n/(2n+4)
1/4+1/3-1/(n+2)>n/(2n+4)
7/12>1/2 恒成立所以原不等式成立~