Given $u_{n+1}=u_n+\frac{1}{n^\alpha u_n}$. Show that $u_n\leqslant \sum_{k=1}^{n}\frac{1}{k^\alpha u_1}$?

Question

Let $u_{n+1}=u_n+\dfrac{1}{n^\alpha u_n}$ for all $n\geqslant1$ and $u_1>0$.

Show that for all $n\geqslant 1$: $$\displaystyle u_n\leqslant \sum_{k=1}^{n}\dfrac{1}{k^\alpha u_1}.$$

I found that :

$$u_{n+1}-u_1\leqslant \sum_{k=1}^{n}\dfrac{1}{k^\alpha u_1},$$

but not the desired result.