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?

The denominator comes from the following:
$$n(a_1^4+...+a_n)^4=(1^2+...+1^2)(a_1^4+_...+a_n^4)\geq(a_1^2+...+a_n^2)^2,$$ which gives
$$a_1^4+...+a_n^4\geq\frac{(a_1^2+...+a_n^2)^2}{n}.$$
Another way: $$\sum_{i=1}^n(a_i^4-1)=\sum_{i=1}^n(a_i-1)(a_i^3+a_i^2+a_i+1)=$$
$$=\sum_{i=1}^n((a_i-1)(a_i^3+a_i^2+a_i+1)-4(a_i-1))=\sum_{i=1}^n(a_i-1)^2(a_i^2+2a_i+3)\geq0.$$
