Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}$

By cauchy-schwarz inequality $$\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{3n+1}\right)(n+1+n+2+\cdots+3n+1)>(1+1+\cdots+1)^2=(2n+1)^2$$ note$$ (n+1+n+2+\cdots+3n+1)=\dfrac{(n+1+3n+1)(2n+1)}{2}=(2n+1)^2$$

You are almost done. Prove that $$\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}-\frac{1}{k+1}$$ is positive.

To do this it is enough to show that $\frac{1}{3k+2}+\frac{1}{3k+4} \gt \frac{2}{3k+3}$. The left side can be written as $\frac{6k+6}{(3k+2)(3k+4}$. So we want to show that $(3k+3)^2\gt (3k+2)(3k+4)$.