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?
 A: Also we can use Jensen. Let $f:\mathbb R \to \mathbb R, x\mapsto x^4$. Then $f$ is convex and thus by Jensen,
$$\frac{a_1^4+\dots+a_n^4}n=\frac{f(a_1)+\dots+f(a_n)}n\geq f\left(\frac{a_1+\dots+a_n}n\right)=\frac{(a_1+\dots+a_n)^4}{n^4}=1.$$
A: I always feel that you want to write things very explicitly and lengthily when using Cauchy-Schwarz, or you may forget square roots, or a factor. Here, we use that for any $(b_i)$,
$$
\left | \sum_{i=1}^n b_i \right |= \left | \sum_{i=1}^n b_i \times 1 \right | \leq \sqrt{\sum_{i=1}^n b_i^2} \sqrt{\sum_{i=1}^n 1^2} = \sqrt{\sum_{i=1}^n b_i^2} \sqrt{n},
$$
then take the square, apply to $b_i = a_i^2$, and then to $b_i = a_i$.
A: 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: $$\sum_{i=1}^n(a_i^4-1)=\sum_{i=1}^n(a_i-1)(a_i^3+a_i^2+a_i+1)=$$
$$=\sum_{i=1}^n((a_i-1)(a_i^3+a_i^2+a_i+1)-4(a_i-1))=\sum_{i=1}^n(a_i-1)^2(a_i^2+2a_i+3)\geq0.$$
A: Note that if $\langle \cdot,\cdot\rangle$ is the scalar product in $\mathbb{R}^{n}$ and $x=(x_{1},...,x_{n}) \in \mathbb{R}^{n}$ is any vector. We have the following:
Let $u$ be the vector full of ones, in other words $u=(1,1,...,1)$
Then by C-S we have
$$
|\langle x,u\rangle|^{2} \leq  \langle u,u\rangle \cdot \langle x,x\rangle
$$
It's easy to see that $\langle u,u\rangle=n$ and $\langle x,u \rangle=x_{1}+x_{2}+...+x_{n}$. Then:
$$
\left(x_{1}+x_{2}+...+x_{n}\right)^{2} \leq n \langle x,x\rangle = n (x_{1}^{2}+...+x_{n}^{2})
$$
In other words:
$$
\frac{\left(x_{1}+x_{2}+...+x_{n}\right)^{2}}{n} \leq x_{1}^{2}+...+x_{n}^{2}
$$
When $x_{i}=a_{i}^{2}$ the result follows.
