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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!

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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): $$ 2n+\sum_{j=1}^n2x_j^2\geq\sqrt{4\left(\sum_{j=1}^n2x_j\right)^2+\left(-2n+\sum_{j=1}^n2x_j^2\right)^2}. $$

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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 $x_1 = x_2 = ...= x_n$

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