Let $\gamma$ be a positive, nondecreasing, continuous, function defined on $[0,\infty]$. Suppose that $\gamma(x+y)\le C(\gamma(x)+\gamma(y))$. In addition, suppose $$ \int_{2}^{\infty}\frac{dr}{\gamma(r)}=\infty.$$
suppose further that 
$$
f(t)\le f(0)+\int_0^t\gamma(f(r))dr+\gamma \Big(\int_0^tf(r)dr\Big).
$$
How to show that 
$$
\gamma\Big(\int_0^t f(r)dr\Big)\le C(t+1)\int_0^t \gamma(f(r))dr
$$

Actually, this question come from [Lemma 1.2 in this paper][1].

**My effort:** Following the hint in this paper, I want to show that $\gamma$ cannot grow faster than $t^2$. However, I can't rule out the possibility that $$\limsup_{t\to +\infty}\frac{\gamma(t)}{t^2}=\infty.$$

I have no idea how to proceed it. Thanks for any help. 


**Update:** If $C\le 1$, I can give a proof.
\begin{align*}
\gamma(\int_0^t f(r)dr)&=\gamma(\sum_{k=1}^n\int_{\frac{(k-1)t}{n}}^{\frac{kt}{n}}f(r)dr)\\
&\le \sum_{k=1}^{n}C^k\gamma(f(\xi_k))\frac{t}{n},\xi_k\in (\frac{k-1}{n}t,\frac{k}{n}t)
\end{align*}
If $C\le 1$, we can send $n$ to $\infty$ and get the proof.

  [1]: https://link.springer.com/article/10.1007%2Fs00205-013-0691-z