How to show $n\sum_{i=1}^n {x_i^2} \ge (\sum_{i=1}^n{x_i})^2$

Question

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.

Answer 1

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 $$\left(\sum_{i=1}^nx_i\right)^2\le\sum_{i=1}^ny_i^2\cdot n$$

Answer 2

The statement is false. Take $x_1 = \dfrac{1}{2}$, and $n = 1$.

Answer 3

This is a special case of Chebyshev's sum inequality for $a_i=b_i$.

Answer 4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ 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 $\ds{\pars{~\mbox{with}\ \lambda \in {\mathbb R}~}}$: $$ 0\leq\pars{\vec{r} + \lambda\vec{\alpha}}^{2} = \alpha^{2}\lambda^{2} +2\vec{r}\cdot\vec{\alpha}\,\lambda + r^{2}\ \imp\ \pars{2\vec{r}\cdot\vec{\alpha}}^{2} - 4\alpha^{2}r^{2}\leq 0\ \imp\ \pars{\vec{r}\cdot\vec{\alpha}}^{2} \leq \alpha^{2}r^{2} $$

Answer 5

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.

$$ \begin{aligned} f\left(x_{1}, x_{2}, \cdots, x_{n}\right) &=n \sum_{i=1}^{n} x_{i}^{2}-\left(\sum_{i=1}^{n} x_{i}\right)^{2} =\sum_{i<j}\left(x_{i}-x_{j}\right)^{2} \ge 0 \end{aligned} $$

$$ \begin{aligned} &\quad \,\,\, n \sum_{i=1}^{n} x_{i}^{2}-\left(\sum_{i=1}^{n} x_{i}\right)^{2} \\ &=n \sum_{i=1}^{n} x_{i}^{2}-\left(\sum_{i=1}^nx_i^2+2\sum_{i<j}x_ix_j\right)\\ &=\sum_{i=1}^n(n-1)x_i^2 - 2\sum_{i<j}x_ix_j \end{aligned} $$

$$ \begin{aligned} &\sum_{i<j} \left(x_i-x_j\right)^2\\ =&\sum_{i<j}\left(x_i^2 -2x_ix_j + x_j^2\right)\\ =&\sum_{i<j}(x_i^2+x_j^2)-2\sum_{i<j}x_ix_j\\ =&\sum_{j=1}^n \sum_{i=1}^{j-1}x_i^2 +\sum_{j=1}^n \sum_{i=1}^{j-1}x_j^2 -2\sum_{i<j}x_ix_j\\ =&\sum_{i=1}^n(n-1)x_i^2 - 2\sum_{i<j}x_ix_j \end{aligned} $$