$x_i = \frac{n(n+1)}{2} + i$ for $i = 1, 2, \ldots, n$

Let $t_i = \frac{n(n+1)}{2} + i$ and $x_i = t_i y_i$ for $i = 1, 2, \ldots, n$. First, note that: $$ \frac{n(n+1)(2n+1)(3n+1)}{12} = \sum_{i = 1}^n it_i $$ Then, the inequality we wished to prove is equivalent to: $$ \sum_{i = 1}^n t_i^2 y_i^2 \geq \frac{1}{n+1} \left(\sum_{i = 1}^n t_i y_i \right)^2 + \left(\sum_{i = 1}^n it_i\right)^{-1} \left(\sum_{i = 1}^n i t_i y_i\right)^2 $$ To begin proving this inequality, observe that by Cauchy-Schwarz inequality, $$ \left(\sum_{i = 1}^n it_i\right)^{-1} \left(\sum_{i = 1}^n i t_i y_i\right)^2 \leq \sum_{i = 1}^n i t_i y_i^2 $$ $$ \left(\sum_{i = 1}^n t_i y_i \right)^2 \leq \left(\sum_{i = 1}^n t_i \right) \left(\sum_{i = 1}^n t_i y_i^2 \right) = \frac{n(n+1)^2}{2} \left(\sum_{i = 1}^n t_i y_i^2 \right) $$ Meanwhile, $$ \sum_{i = 1}^n t_i^2 y_i^2 = \sum_{i = 1}^n \left( \frac{n(n+1)}{2} + i \right) t_i y_i^2 = \frac{n(n+1)}{2} \left(\sum_{i = 1}^n t_i y_i^2 \right) + \sum_{i = 1}^n i t_i y_i^2 $$ We are done.

Since the expression is really homogenous in terms of all variables, look at the equality cases first. It helps when you look at some equality cases for small $n$.

IndoMathXdZ wrote: Fantastic Inequality! Here's a little motivation behind the following solution. The hard part for me in this problem is in fact to find when the equality holds. The rest quickly follows. The equality holds when $x_i = \frac{n(n+1)}{2} + i$. First, notice that \[ \left( \sum_{i = 1}^n i \right)^2 + \sum_{i = 1}^n i^2 = \frac{n(n+1)(n+2)(3n+1)}{12}\]Now, notice that we could let $x_i = S + a_i$ for $i \in \mathbb{N}$, where $S = a_1 + a_2 + \dots + a_n$. Therefore, we wanted to prove \[ \sum_{i = 1}^n x_i^2 \ge \frac{1}{n+1} \left( \sum_{i = 1}^n x_i \right)^2 + \frac{12 \left( \sum_{i = 1}^n i x_i \right)^2 }{n(n+1)(n+2)(3n+1)} \]which is equivalent to \[ \sum_{i = 1}^n (S + a_i)^2 \ge \frac{1}{n+1} \left( \sum_{i = 1}^n (S + a_i) \right)^2 + \frac{\left( \sum_{i = 1}^n iS + ia_i \right)^2 }{ \left( \sum_{i = 1}^n i \right)^2 + \sum_{i = 1}^n i^2 } \]\[ \left( S^2 + \sum_{i = 1}^n a_i^2 \right) \left( \left( \sum_{i = 1}^n i \right)^2 + \sum_{i = 1}^n i^2 \right) \ge \left( \sum_{i = 1}^n iS + \sum_{i = 1}^n ia_i \right)^2 \]which is indeed true by Cauchy Schwarz.

Define the three vectors \begin{align*} \mathbf{x} & =\left( x_{1},x_{2},\ldots,x_{n}\right) ^{T},\\ \mathbf{a} & =\left( 1,1,\ldots,1\right) ^{T},\\ \mathbf{b} & =\left( 1,2,\ldots,n\right) ^{T} \end{align*}in the Euclidean space $\mathbb{R}^{n}$. Let $\left\vert \left\vert \mathbf{v}\right\vert \right\vert $ denote the norm of a vector $\mathbf{v}=\left( v_{1},v_{2},\ldots,v_{n}\right) ^{T}\in\mathbb{R}^{n}$; this is defined by $\left\vert \left\vert \mathbf{v}\right\vert \right\vert =\sqrt{\mathbf{v}^{T}\mathbf{v}}=\sqrt{v_{1}^{2}+v_{2}^{2}+\cdots+v_{n}^{2}}$. (It is also known as the Euclidean length of $\mathbf{v}$.) Let \[ \alpha=\dfrac{1}{n+1}\qquad\text{and}\qquad\beta=\dfrac{12}{n\left( n+1\right) \left( n+2\right) \left( 3n+1\right) }. \]Now, the inequality in question rewrites as \[ \left\vert \left\vert \mathbf{x}\right\vert \right\vert ^{2}\geq\alpha\left( \mathbf{a}^{T}\mathbf{x}\right) ^{2}+\beta\left( \mathbf{b}^{T} \mathbf{x}\right) ^{2}. \]How to prove an inequality like this? We first define $U$ to be the span of the two vectors $\mathbf{a}$ and $\mathbf{b}$ in $\mathbb{R}^{n}$, and we let $\mathbf{y}$ be the orthogonal projection of $\mathbf{x}$ onto $U$. Then, $\left\vert \left\vert \mathbf{y}\right\vert \right\vert \leq\left\vert \left\vert \mathbf{x}\right\vert \right\vert $ (since an orthogonal projection of a vector is always at most as long as the vector itself), but $\mathbf{a}^{T}\mathbf{y}=\mathbf{a}^{T}\mathbf{x}$ (since $\mathbf{x} -\mathbf{y}$ is orthogonal to $U$ and thus also to $\mathbf{a}$) and $\mathbf{b}^{T}\mathbf{y}=\mathbf{b}^{T}\mathbf{x}$ (likewise). Hence, in order to prove the inequality \[ \left\vert \left\vert \mathbf{x}\right\vert \right\vert ^{2}\geq\alpha\left( \mathbf{a}^{T}\mathbf{x}\right) ^{2}+\beta\left( \mathbf{b}^{T} \mathbf{x}\right) ^{2}, \]it suffices to prove the at-least-as-strong inequality \[ \left\vert \left\vert \mathbf{y}\right\vert \right\vert ^{2}\geq\alpha\left( \mathbf{a}^{T}\mathbf{y}\right) ^{2}+\beta\left( \mathbf{b}^{T} \mathbf{y}\right) ^{2}. \]But the latter inequality is easy, since $\mathbf{y}$ is stuck in a low-dimensional subspace: Namely, we have $\mathbf{y}\in U$, so that $\mathbf{y}$ is a linear combination of $\mathbf{a}$ and $\mathbf{b}$; that is, \[ \mathbf{y}=\left( c+1d,c+2d,\ldots,c+nd\right) ^{T} \]for some $c,d\in\mathbb{R}$. Using these $c$ and $d$, we can compute $\left\vert \left\vert \mathbf{y}\right\vert \right\vert ^{2}$, $\mathbf{a}^{T}\mathbf{y}$ and $\mathbf{b}^{T}\mathbf{y}$ directly: \begin{align*} \left\vert \left\vert \mathbf{y}\right\vert \right\vert ^{2} & =\left( c+1d\right) ^{2}+\left( c+2d\right) ^{2}+\cdots+\left( c+nd\right) ^{2}\\ & =nc^{2}+2c\underbrace{\left( 1+2+\cdots+n\right) }_{=\dfrac{n\left( n+1\right) }{2}}d+\underbrace{\left( 1^{2}+2^{2}+\cdots+n^{2}\right) }_{=\dfrac{n\left( n+1\right) \left( 2n+1\right) }{6}}d^{2}\\ & =nc^{2}+n\left( n+1\right) cd+\dfrac{n\left( n+1\right) \left( 2n+1\right) }{6}d^{2}; \end{align*}\begin{align*} \mathbf{a}^{T}\mathbf{y} & =\left( c+1d\right) +\left( c+2d\right) +\cdots+\left( c+nd\right) \\ & =nc+\underbrace{\left( 1+2+\cdots+n\right) }_{=\dfrac{n\left( n+1\right) }{2}}d=nc+\dfrac{n\left( n+1\right) }{2}d; \end{align*}\begin{align*} \mathbf{b}^{T}\mathbf{y} & =1\left( c+1d\right) +2\left( c+2d\right) +\cdots+n\left( c+nd\right) \\ & =\underbrace{\left( 1+2+\cdots+n\right) }_{=\dfrac{n\left( n+1\right) }{2}}c+\underbrace{\left( 1^{2}+2^{2}+\cdots+n^{2}\right) }_{=\dfrac {n\left( n+1\right) \left( 2n+1\right) }{6}}d\\ & =\dfrac{n\left( n+1\right) }{2}c+\dfrac{n\left( n+1\right) \left( 2n+1\right) }{6}d. \end{align*}Hence, the inequality in question (that is, $\left\vert \left\vert \mathbf{y}\right\vert \right\vert ^{2}\geq\alpha\left( \mathbf{a}^{T}\mathbf{y}\right) ^{2}+ \beta\left( \mathbf{b}^{T}\mathbf{y}\right)^{2}$) rewrites as \begin{align*} & nc^{2}+n\left( n+1\right) cd+\dfrac{n\left( n+1\right) \left( 2n+1\right) }{6}d^{2}\\ & \geq\alpha\left( nc+\dfrac{n\left( n+1\right) }{2}d\right) ^{2} +\beta\left( \dfrac{n\left( n+1\right) }{2}c+\dfrac{n\left( n+1\right) \left( 2n+1\right) }{6}d\right) ^{2}. \end{align*}After long and painful computations, this inequality boils down to \[ n\left( n-1\right) \left( dn^{2}+dn-2c\right) ^{2}\geq0, \]which is clearly true.