inequality - For any natural number $n$, Prove that $prod^{n}_{r=1}bigg(r+frac{1}{n}bigg)leq 2(n!)$

For any natural number $n$, Prove that $$\displaystyle \prod^{n}_{r=1}\bigg(r+\frac{1}{n}\bigg)\leq 2(n!)$$

Trial Solution: Using $\displaystyle \frac{1}{n}\leq 1,2,3,\cdots n$

$\displaystyle \prod^{n}_{r=1}\bigg(1+\frac{2}{n}\bigg)\leq 2\cdot 4\cdot 6\cdots \cdots 2n$

$$\prod^{n}_{r=1}\bigg(1+\frac{2}{n}\bigg)\leq 2^n\cdot n!$$

Could some help me how to prove my original inequality, Thanks

Answer

We can verify directly for $n=1,2$. Suppose $n\geq3$. Then
$$\sum_{r=1}^n\log\left(1+\frac1{rn}\right) \leq\sum_{r=1}^n\frac1{rn} \leq\frac{1+1/2+(n-2)/3}{n} \leq\frac{11}{18}<\log 2.$$
Now exponentiate and multiply by $n!$.