Prove that there exists infinitely many positive integers $n$,such $nS_{p}(n)+2014n\ge S_{p}(n^2)+2014S_{p}(n)$ 
Prove that there exists infinitely many positive integers $n$
such
$$nS_{p}(n)+2014S_{p}(n)\ge S_{p}(n^2)+2014n$$
or
$$(n+2014)S_{p}(n)\ge S_{p}(n^2)+2014\cdot n$$

Here $S_{p}(n)$ is the sum of the digits of $n$ when written in base $p$,where $p$ is prime
if $n$ is equal to $p^k$,we have
$$nS_{p}(n)+2014S_{p}(n)=n+2014$$
$$S_{p}(n^2)+2014n=2014n +1$$ this case  is  not true for $n=p^k$
 A: This proof does not require that $p$ is prime (it needs only $p$ to be an integer greater than $1$).  Let $\mathbf{r}:=\left\{r_i\right\}_{i=0}^k$ be a sequence of nonnegative integers such that $r_i>2\,r_{i-1}$ for all $i=1,2,\ldots,k$, and define $n(\mathbf{r})$ to be $\sum_{i=0}^k \,p^{r_i}$.  Note that $S_p\big(n(\mathbf{r})\big)=k+1$ and
$$\big(n(\mathbf{r})\big)^2=\sum_{i=0}^k\,\sum_{j=0}^k\,p^{r_i+r_j}\,,$$
so $S_p\Big(\big(n(\mathbf{r})\big)^2\Big)\leq \Big(S_p\big(n(\mathbf{r})\big)\Big)^2=(k+1)^2$.  That is, $$\big(n(\mathbf{r})+2014\big)\,S_p\big(n(\mathbf{r})\big)=(k+1)\big(n(\mathbf{r})+2014\big)\,,$$
while
$$\Big(S_p\big(n(\mathbf{r})\big)\Big)^2+2014\,n(\mathbf{r})\leq (k+1)^2+2014\,n(\mathbf{r})\,.$$
If $k\geq 2014$, then $n(\mathbf{r})\geq k+1$, $(k+1)^2-2014(k+1)=(k-2013)(k+1)$, and $(k+1)\,n(\mathbf{r})-2014\,n(\mathbf{r})=(k-2013)\,n(\mathbf{r})$.  Ergo,
$$(k+1)^2-2014(k+1) \leq (k+1)\,n(\mathbf{r})-2014\,n(\mathbf{r})\,.$$
Consequently,
$$
\begin{align}
\Big(S_p\big(n(\mathbf{r})\big)\Big)^2+2014\,n(\mathbf{r})
&\leq (k+1)^2+2014\,n(\mathbf{r}) 
\\
&\leq (k+1)\big(n(\mathbf{r})+2014\big) 
\\
&= \big(n(\mathbf{r})+2014\big)\,S_p\big(n(\mathbf{r})\big)\,.
\end{align}$$
Hence, all integers of the form $n(\mathbf{r})$ satisfy the required inequality.  One can also replace $2014$ by any $N\in\mathbb{N}$ (in which case, $k$ is taken to be at least $N$).
