Let 


* $R = \mbox{diag} (r\_1,\dots,r\_n)$, where $r\_1, \dots, r\_n > 0$, be a (positive) diagonal matrix.
* $1\_n \in \mathbb{R}^n $ denote the $n$-dimensional vector of all ones.
* $S$ be a matrix defined by $[S]\_{jk}=\sin (\theta\_{j}-\theta\_k +\delta)$, where $[\theta\_1, \dots, \theta\_n]^\mathrm{T} \in \mathbb{R}^n$ and $\delta \in \mathbb R$.


Is it true that $$\left( 1^\mathrm{T}\_n S^T R S 1\_n -1^\mathrm{T}\_n R S R^{-1} S 1\_n \right) \geq 0$$ always holds? 




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My thoughts so far:


Since $S^T R S = (R^{1/2} S)^\mathrm{T} (R^{1/2}S)$, therefore it is positive semidefinite and so $1^\mathrm{T}\_n S^T R S 1\_n \geq 0$. 


Also, I see that when $1^\mathrm{T}\_n S^T R S 1\_n = 0$, then $1^\mathrm{T}\_n R S R^{-1} S 1\_n =0$.



Is this expression always non-negative?
Consider $R = {\rm diag}(r, 1/r)$ and $\delta = \pi/4, \theta\_1 = \pi/4, \theta\_2 = 0$. Then $$
S = \left[\begin{array}{cc} \sin(\theta\_1 - \theta\_1 + \delta) & \sin(\theta\_1 - \theta\_2 + \delta) \\ \sin(\theta\_2 - \theta\_1 + \delta) & \sin(\theta\_2 - \theta\_2 + \delta) \end{array}\right] = 
\left[ \begin{array}{cc} \sin(\pi/4) & \sin(\pi/2) \\ \sin(0) & \sin(\pi/4) \end{array}\right]= \left[\begin{matrix} 1/\sqrt 2 & 1 \\ 0 & 1/\sqrt 2\end{matrix}\right].$$ 
Now
$$
S^TRS = \left[\begin{matrix} r/\sqrt 2 & 0 \\ r & 1/(\sqrt 2 r)\end{matrix}\right]\left[\begin{matrix} 1/\sqrt 2 & 1 \\ 0 & 1/\sqrt 2\end{matrix}\right] = \left[\begin{array}{cc} r/2 & r/\sqrt 2 \\ r/\sqrt 2 & r + 1/(2r) \end{array}\right] 
$$
$$ RSR^{1/2}S = \left[\begin{matrix} 1/\sqrt 2 & r^2 \\ 0 & 1/\sqrt 2\end{matrix}\right]\left[\begin{matrix} 1/\sqrt 2 & 1 \\ 0 & 1/\sqrt 2\end{matrix}\right]= \left[\begin{array}{cc} 1/2 & (1+r^2)/\sqrt 2 \\ 0 & 1/2 \end{array}\right]
$$
and so
$$
1\_2^TS^TRS1\_2 = (3/2 + \sqrt 2)r + 1/(2r) \ \ \textrm{and} \ \ 1\_2^TRSR^{-1}S1\_2 = 1 + 1/\sqrt 2 + r^2/\sqrt 2.
$$
Therefore, $$\lim\_{r\rightarrow \infty} 1\_2^TS^TRS1\_2 - 1\_2^TRSR^{-1}S1\_2 = -\infty$$
and so your desired inequality does not hold in general.


