Inequality with summations and roots I'm trying to prove that
$$\frac{1}{2}(2n + \sum_{i=1}^{n}2x_i^2-\sqrt{4(\sum_{i=1}^{n}2x_i)^2 + (-2n + \sum_{i=1}^{n}2x_i^2)^2}) > 0$$
For all $x_i \in {\rm I\!R}$ and $n>0$ I tried using the fact that
$$(\sum_{i=1}^{n}2x_i)^2 = (-\sum_{i=1}^{n}2x_i)^2$$
and then invoking the identity
$$\sqrt{a_1 + a_2} \leq \sqrt{a_1} + \sqrt{a_2}$$
However, I only simplified to
$$2(n + \sum_{i=1}^{n}x_i) > 0$$
which can be negative in case $$\sum_{i=1}^{n}x_i<-n<0$$
Any advice and help is highly appreciated. Thank you!
 A: Let $a = \displaystyle \sum_{k=1}^n x_k, b = \displaystyle \sum_{k=1}^n x_k^2\implies n+b-\sqrt{4a^2+(b-n)^2}>0\iff (n+b)^2 > 4a^2+(b-n)^2\iff 4bn > 4a^2\iff bn > a^2$, which is true due to Cauchy-Schwarz inequality, and $=$ occurs when $x_1 = x_2 = ...= x_n$
A: From Cauchy-Schwarz,
$$
\left(\sum_{j=1}^nx_j\right)^2\leq n\,\sum_{j=1}^nx_j^2.
$$
Multiplying by 4 and splitting the sum,
$$
2n\sum_{j=1}^nx_j^2\geq4\left(\sum_{j=1}^nx_j\right)^2-2n\sum_{j=1}^nx_j^2.
$$
Adding terms on both sides, 
$$
n^2+2n\sum_{j=1}^nx_j^2+\left(\sum_{j=1}^nx_j^2\right)^2\geq4\left(\sum_{j=1}^nx_j\right)^2+n^2-2n\sum_{j=1}^nx_j^2+\left(\sum_{j=1}^nx_j^2\right)^2.
$$
Compacting the binomial expressions,
$$
\left(n+\sum_{j=1}^nx_j^2\right)^2\geq4\left(\sum_{j=1}^nx_j\right)^2+\left(-n+\sum_{j=1}^nx_j^2\right)^2.
$$
Multiply both sides by 4:
$$
\left(2n+\sum_{j=1}^n2x_j^2\right)^2\geq4\left(\sum_{j=1}^n2x_j\right)^2+\left(-2n+\sum_{j=1}^n2x_j^2\right)^2.
$$
Take square root (both sides of the inequality are positive):
$$
2n+\sum_{j=1}^n2x_j^2\geq\sqrt{4\left(\sum_{j=1}^n2x_j\right)^2+\left(-2n+\sum_{j=1}^n2x_j^2\right)^2}.
$$
