er $i,j$ will take values from $\{1,\dots,n\}$ and $k,l,m$ will take values from $\{1,\dots,r\}$. We simplify the above as $$\biggl(\sum_i c_i\biggr)\biggl(\sum_j\sum_{\substack{{k,l}\\k\neq l}}x_{jk}x_{jl}\biggr)
\leq c\biggl(\sum_{i,j}\sum_{\substack{{k,l}\\k\neq
l}}x_{ik}x_{jl}\biggr),$$ then we factor out and also utilize the symmetry in $k,l$ to arrive at the equivalent form $$\sum_{i,j}c_i\sum_{\substack{{k,l}\\k<l}}x_{jk}x_{jl}\leq
\sum_{i,j}c\sum_{\substack{{k,l}\\k<l}}x_{ik}x_{jl}.$$ We distribute the terms in $i,j$ on both sides as follows: $$\sum_i c_i\sum_{\substack{{k,l}\\k<l}}x_{ik}x_{il}+
\sum_{\substack{{i,j}\\i<j}}\biggl(c_i\sum_{\substack{{k,l}\\k<l}}x_{jk}x_{jl}
+c_j\sum_{\substack{{k,l}\\k<l}}x_{ik}x_{il}\biggr)\leq \sum_i
c\sum_{\substack{{k,l}\\k<l}}x_{ik}x_{il}+
\sum_{\substack{{i,j}\\i<j}}c\sum_{\substack{{k,l}\\k<l}}(x_{ik}x_{jl}+x_{jk}x_{il}).$$ It is clear that $$c_i\sum_{\substack{{k,l}\\k<l}}x_{ik}x_{il}\leq c\sum_{\substack{{k,l}\\k<l}}x_{ik}x_{il},\quad 1\leq i\leq n,$$ therefore it suffices to show that $$c_i\sum_{\substack{{k,l}\\k<l}}x_{jk}x_{jl}+c_j\sum_{\substack{{k,l}\\k<l}}x_{ik}x_{il}
\leq c\sum_{\substack{{k,l}\\k<l}}(x_{ik}x_{jl}+x_{jk}x_{il}),\quad
1\leq i<j\leq n.$$ We will prove this in the stronger form $$c_i\sum_{\substack{{k,l}\\k<l}}x_{jk}x_{jl}+c_j\sum_{\substack{{k,l}\\k<l}}x_{ik}x_{il}
\leq
c_i\sum_{\substack{{k,l}\\k<l}}(x_{ik}x_{jl}+x_{jk}x_{il}),\quad
1\leq i<j\leq n.$$

We now fix $1\leq i<j\leq n$ and introduce $x_k:=x_{ik}$, $x'_k:=x_{jk}$. Then the previous inequality reads $$\biggl(\sum_m
x_m\biggr)\biggl(\sum_{\substack{{k,l}\\k<l}}x'_kx'_l\biggr)+
\biggl(\sum_m
x'_m\biggr)\biggl(\sum_{\substack{{k,l}\\k<l}}x_kx_l\biggr)\leq
\biggl(\sum_m
x_m\biggr)\sum_{\substack{{k,l}\\k<l}}(x_kx'_l+x'_kx_l),$$ that is, $$\sum_{\substack{{k,l,m}\\k<l}}(x_mx'_kx'_l+x_kx_lx'_m)\leq
\sum_{\substack{{k,l,m}\\k<l}}(x_kx_mx'_l+x_lx_mx'_k).$$ The right hand side equals $$\begin{aligned}
\sum_{\substack{{k,l,m}\\k<l}}(x_kx_mx'_l+x_lx_mx'_k)
&=\sum_{\substack{{k,l,m}\\l\neq
k}}