Xét $f(x)=\arctan x,x \in [n,n+1]$
Theo định lý Lagrange:
$\exists c \in (n,n+1)$
$f(n+1)-f(n)=f'(c)(n+1-n)$
$\Rightarrow \arctan (n+1)-\arctan n=\frac{1}{1+c^{2}}$
Mà: $\arctan a-\arctan b=\arctan (\frac{a-b}{1+ab})(1)$
$\Rightarrow \arctan \frac{1}{n^{2}+n+1}=\frac{1}{1+c^{2}}$
Hơn nữa: $n<c<n+1$
$\Leftrightarrow n^{2}<c^{2}<(n+1)^{2}$
$\Leftrightarrow 1+n^{2}<1+c^{2}<1+(n+1)^{2}$
$\Leftrightarrow \frac{1}{1+(n+1)^{2}}<\frac{1}{1+c^{2}}<\frac{1}{1+n^{2}}$
Vậy: $\frac{1}{1+(n+1)^{2}}<\arctan \frac{1}{n^{2}+n+1}<\frac{1}{1+n^{2}}$
$\Rightarrow $ (ĐPCM)
(Chứng minh
(1):
Ta có :
$\tan(\arctan a-\arctan b)=\frac{\tan(\arctan a)-\tan(\arctan b)}{1+\tan(\arctan
a).\tan(\arctan b)}=\frac{a-b}{1+ab}$
$\Rightarrow\arctan
a-\arctan b=\arctan(\frac{a-b}{1+ab})$)