How to show $n\sum_{i=1}^n {x_i^2} \ge (\sum_{i=1}^n{x_i})^2$ How can I show that $n\sum_{i=1}^n {x_i^2} \ge (\sum_{i=1}^n{x_i})^2$ for any natural number $n$ and $x_i \in\mathbb{R}?$ I assume there is something about Cauchy-Schwarz and induction, but I really don't see it.
 A: Assuming $x_i\ge0$ for each $i$, $x_i\le\sum_{k=1}^nx_k$ and $x_i^2\le(\sum_{k=1}^nx_k)^2$ for each $i$, so
$$\sum_{i=1}^nx_i^2\le\sum_{i=1}^n\left(\sum_{k=1}^nx_k\right)^2=n\left(\sum_{k=1}^nx_k\right)^2$$
The updated version follows from the Cauchy-Schwarz inequality, with $y_i=1$ in:
$$\left(\sum_{i=1}^nx_iy_i\right)^2\le\sum_{i=1}^nx_i^2\sum_{i=1}^ny_i^2$$
to yield 
$$\left(\sum_{i=1}^nx_i\right)^2\le\sum_{i=1}^ny_i^2\cdot n$$
A: $\newcommand{\+}{^{\dagger}}
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Lets $\ds{\vec{\alpha} = \overbrace{\pars{1,1,\ldots,1}}^{\ds{n\,\,\,\,\, 1'\mbox{s}}}\ \mbox{and}\ \vec{r}\equiv\pars{x_{1},x_{2},\ldots,x_{n}}}$:

\begin{align}
\pars{\vec{r}\cdot\vec{\alpha}}^{2} &\leq r^{2}\alpha^{2}\tag{1}
\\[3mm]\vec{r}\cdot\vec{\alpha} =\sum_{i = 1}^{n}x_{i}\,,\qquad
r^{2} &= \sum_{i = 1}^{n}x_{i}^{2}\,,\qquad \alpha^{2} = \sum_{i = 1}^{n}1^{2}=n
\end{align}

$$\color{#00f}{\large%
n\sum_{i = 1}^{n}x_{i}^{2} \geq \pars{\sum_{i = 1}^{n}x_{i}}^{2}}
$$

$\pars{1}$ arises from $\ds{\pars{~\mbox{with}\ \lambda \in {\mathbb R}~}}$:
  $$
0\leq\pars{\vec{r} + \lambda\vec{\alpha}}^{2} = \alpha^{2}\lambda^{2}
+2\vec{r}\cdot\vec{\alpha}\,\lambda + r^{2}\
\imp\ \pars{2\vec{r}\cdot\vec{\alpha}}^{2} - 4\alpha^{2}r^{2}\leq 0\
\imp\ \pars{\vec{r}\cdot\vec{\alpha}}^{2} \leq \alpha^{2}r^{2}
$$

A: The statement is false. Take $x_1 = \dfrac{1}{2}$, and $n = 1$.
A: This is a special case of Chebyshev's sum inequality for $a_i=b_i$.
A: This question is equivalent to proving that the following quadratic form is positive semidefinite：
$$
f(x_1,x_2,\cdots,x_n) = n\sum_{i=1}^nx_i^2-\left(\sum_{i=1}^n x_i\right)^2
$$
if you can notice the following equation, then we can easily proof the answer.
$$
\begin{aligned}
f\left(x_{1}, x_{2}, \cdots, x_{n}\right) &=n \sum_{i=1}^{n} x_{i}^{2}-\left(\sum_{i=1}^{n} x_{i}\right)^{2} =\sum_{i<j}\left(x_{i}-x_{j}\right)^{2} \ge 0
\end{aligned}
$$
$$
\begin{aligned}
&\quad  \,\,\, n \sum_{i=1}^{n} x_{i}^{2}-\left(\sum_{i=1}^{n} x_{i}\right)^{2} \\
&=n \sum_{i=1}^{n} x_{i}^{2}-\left(\sum_{i=1}^nx_i^2+2\sum_{i<j}x_ix_j\right)\\
&=\sum_{i=1}^n(n-1)x_i^2 - 2\sum_{i<j}x_ix_j
\end{aligned}
$$
$$
\begin{aligned}
&\sum_{i<j} \left(x_i-x_j\right)^2\\
=&\sum_{i<j}\left(x_i^2 -2x_ix_j + x_j^2\right)\\
=&\sum_{i<j}(x_i^2+x_j^2)-2\sum_{i<j}x_ix_j\\
=&\sum_{j=1}^n \sum_{i=1}^{j-1}x_i^2 +\sum_{j=1}^n \sum_{i=1}^{j-1}x_j^2 -2\sum_{i<j}x_ix_j\\
=&\sum_{i=1}^n(n-1)x_i^2 - 2\sum_{i<j}x_ix_j 
\end{aligned}
$$
