<s>[INST] Inequality with $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}$ Inspired by this recent question, I suggest this.
Let $n=2,3,4, \ldots .$ Then
$$
\frac{7}{12} < \cfrac 1 {1 + \cfrac {1^2} {1 + \cfrac {2^2} {\ddots + \cfrac \vdots { 1 + \, {n^2} \,}}}}  \leq \frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} \tag1
$$
Could you prove $(1)$?
 [/INST] Here is a brute force method to prove the first inequality. We will see the calculations for $n=2,3,4,5$ and that will be sufficient to prove for general $n$. Here goes: 
$n=2$
$$\frac{7}{12} \leq \cfrac{1}{1+\cfrac{1^2}{1 + 2^2}} \iff \frac{12}{7} \geq 1+\cfrac{1^2}{1 + 2^2} \iff \frac{5}{7} \geq \cfrac{1^2}{1 + 2^2}$$
which is true. For $n=3$, we'll take the calculations from here to get
$$ \frac{5}{7} \geq \cfrac{1^2}{1 + \cfrac{2^2}{1+3^2}} \iff \frac{7}{5} \leq 1 + \cfrac{2^2}{1+3^2} \iff \frac{2}{5} \leq \cfrac{2^2}{1+3^2} $$ which is again true. Carrying on for $n=4$
$$ \frac{2}{5} \leq \cfrac{2^2}{1+ \cfrac{3^2}{1+4^2}} \iff 10 \geq 1+ \cfrac{3^2}{1+4^2} \iff 1 \geq \cfrac{1}{1+4^2} $$ which is true. Carrying on for $n=5$, we have
$$ 1 \geq \cfrac{1}{1+ \cfrac{4^2}{1+5^2}} \iff 1 \leq 1+ \frac{4^2}{1+5^2}  \iff 0 \leq \frac{4^2}{1+5^2} $$ which is true.
For $n\geq 6$, the equivalent condition will just be 
$$ 0 \leq \cfrac{4^2}{1+ \cfrac{5^2}{1+ \cfrac{6^2}{1+ ...}}} $$
which is of course true. These calculations are not revealing, but I sometimes find them cute. 
</s>