Inequality on a sequence of bounded reals

Let $a_j$ be a sequence of positive reals. Show that, $$\left ( \sum_{j=1}^N a_j \right)^{\theta} \le N^{\theta-1} \sum_{j=1}^N a_j^{\theta}$$ when $1 \le \theta < \infty$.

$\textbf{My Attempt:}$ I have been playing around with this inequality for some time but most of my attempts have been fruitless.

I can prove that $$\left ( \sum_{j=1}^N a_j \right)^{\theta} \ge \sum_{j=1}^N a_j^{\theta}$$ when $1 \le \theta < \infty$ using an integration trick, however, I do not know how to show such a constant exists.

$$\sum_{j = 1}^N a_j \le \left(\sum_{j = 1}^N a_j^\theta\right)^{\frac{1}{\theta}} \left(\sum_{j = 1}^N 1^{\frac{\theta}{\theta - 1}}\right)^{\frac{\theta - 1}{\theta}} = N^{\frac{\theta - 1}{\theta}} \left(\sum_{j = 1}^N a_j^\theta\right)^{\frac{1}{\theta}}.$$
$$\left(\sum_{j = 1}^N a_j\right)^\theta \le N^{\theta - 1} \sum_{j = 1}^N a_j^\theta$$
This follows from the fact that the map $$(a_1,a_2,\ldots,a_n)\mapsto a_1^{\theta}+a_2^{\theta}+\ldots+a_n^{\theta}$$ is convex - that is, if you average two vector inputs, you get a smaller value out. This is easily proven since it is a sum of maps of the form $(a_1,a_2,\ldots,a_n)\mapsto a_i^{\theta}$, each of which is convex. In particular, if we restrict our attention to where we fix $a_1+a_2+\ldots +a_n=C$ - that is, fix the left hand side. Then, we notice that the vectors $(a_1,a_2,\ldots,a_n)$, $(a_2,a_3,\ldots,a_n,a_1)$, $(a_3,a_4,\ldots,a_n,a_1,a_2)$, $\ldots$, $(a_n,a_1,\ldots,a_{n-2},a_{n-1})$ all map to the same value, and their average is of the form $(\alpha,\alpha,\ldots,\alpha)$ for some $\alpha$ (but by convexity this maps to a lesser value than the original vectors), it follows that, by averaging all the inputs, we lessen the right hand side while leaving untouched the left hand side. Thus it suffices to prove: $$\left(\sum_{j=1}^N \alpha\right)^{\theta}\leq N^{\theta-1}\left(\sum_{j=1}^N \alpha^{\theta}\right)$$ which is obvious.