已知函数f(x)=ln(x+a)－x^2－x在x＝0处取得极值.

f(x)=ln(x+a)-x^2-x 显然a＞0
f'(x)=1/(x+a)-2x-1
∵f(x)在点x=0处取极值,
∴f'(0)=0,可得a=1
∴f(x)=ln(1+x)-x^2-x
f'(x)=-x(x+3)/(x+1)
当-1＜x＜0时f'(x)＞0,当x＞0时f'(x)＜0,∴在点x=0处f(x)取极大值-2,当x＞0时,ln(1+x)＜x^2+x.
取x=1/n (n∈N*）,可得(n+1)/(n^2)＞ln[(n+1)/n],再取n=1,2,3,…,n,将所得的n个式子相加,就可得出所要证的式子.