Cauchy-Schwarz inequality for $a_1^4 + a_2^4 + \cdots + a_n^4 \geqslant n$ 
Let $a_1+a_2,...,a_n \in \mathbb{R}.$ Show that if $a_1+a_2+...+a_n=n$, then $$a_1^4+a_2^4+...+a_n^4 \geqslant n.$$

The proposed solution for this was the following:
Using the Cauchy-Schwarz inequality twice we get
$a_1^4+a_2^4+...+a_n^4 \geqslant \frac{(a_1^2+a_2^2+...+a_n^2)^2}{n} \geqslant \frac{\frac{((a_1+a_2+...+a_n)^2)^2}{n}}{n} = \frac{(\frac{n^2}{n})^2}{n} = n$
I can see that we can deduce this straight from the definition, but where on earth does the denominator $n$ come for $a_1^4+a_2^4+...+a_n^4 \geqslant \frac{(a_1^2+a_2^2+...+a_n^2)^2}{n}$.
From Cauchy-Schwarz we can come up with $a_1^4+a_2^4+...+a_n^4 \geqslant (a_1^2+a_2^2+...+a_n^2)^2$, but I don't see where the denominator comes from. Could someone enlighten me?

Also we can use Jensen. Let $f:\mathbb R \to \mathbb R, x\mapsto x^4$. Then $f$ is convex and thus by Jensen,
$$\frac{a_1^4+\dots+a_n^4}n=\frac{f(a_1)+\dots+f(a_n)}n\geq f\left(\frac{a_1+\dots+a_n}n\right)=\frac{(a_1+\dots+a_n)^4}{n^4}=1.$$
