Inequality on singular values of two matrices Let $\Sigma_1=\begin{pmatrix} \sigma_1^1 & 0\\0 &\sigma_2^1 \end{pmatrix}$ and $A=R^{-1}\Sigma_2R$, $\Sigma_2=\begin{pmatrix} \sigma_1^2 & 0\\0 &\sigma_2^2 \end{pmatrix}$, $R=\begin{pmatrix} \cos\theta & \sin\theta\\-\sin\theta &\cos\theta \end{pmatrix}$.
Suppose $\sigma_1^1\ge \sigma_2^1,\sigma_1^2,\sigma_2^2\ge 0$ and fix any $\theta$.
This is my question:
Is it true that, for every $v_1=(x_1,y_1),v_2=(x_2,y_2)\in \mathbb{R}^2$ such that all four coordinates $x_1,x_2,y_1,y_2$ are $\ge 0$, the following inequality is satisfied?
$$\frac{|\Sigma_1v_1+Av_2|}{|v_1+v_2|}\le \sigma_1^1$$

Edit: I'm quite sure that this inequality wouldn't be satisfied in case of two generic vectors $v_1,v_2\in \mathbb{R}^2$. Just consider $v_1=(1,0)$ and $v_2=(-1/2,1/2)$ and $A$ such that $A(v_2)=(1/2,1/2)$.
 A: This is not true. Let $v_1=(1,0)^T,\ v_2=(0,1)^T,\ \Sigma_1=I$ and
$$
A=\pmatrix{\cos\theta\\ -\sin\theta}\pmatrix{\cos\theta&-\sin\theta}=\pmatrix{\cos^2\theta&-\sin\theta\cos\theta\\ -\sin\theta\cos\theta&\sin^2\theta}.
$$
Then
$$
\|\Sigma_1v_1+Av_2\|=\left\|\pmatrix{1-\sin\theta\cos\theta\\ \sin^2\theta}\right\|>\sqrt{2}=\sigma_1^1\|v_1+v_2\|
$$
when $\theta$ is slightly greater than $\frac\pi2$.
A: [Hint]
The eigenvalues of $\Sigma_1$ are $\sigma_1^1$ and $\sigma_2^1$; the eigenvectores are $e_1^1 = \{1,0\}^T$ and $e_2^1 = \{0,1\}^T$.
The eigenvalues of $A$ are $\sigma_1^2$ and $\sigma_2^2$; the eigenvectores are $e_1^2 = \{cos\,\theta,sin\,\theta\}^T$ and $e_2^2 = \{-sin\,\theta,cos\,\theta\}^T$.
$\Sigma_1\,v_1+A\,v_2 = (v_1\cdot e_1^1)\,\Sigma_1\,e_1^1 + (v_1\cdot e_2^1)\,\Sigma_1\,e_2^1 + (v_2\cdot e_1^2)\,A\,e_1^2 + (v_2\cdot e_2^2)\,A\,e_2^2 = \\
x_1\,\sigma_1^1\,e_1^1+y_1\,\sigma_2^1\,e_2^1 + (x_2\,cos\,\theta+y_2\,sin\,\theta)\,\sigma_1^2\,e_1^2 + (-x_2\,sin\,\theta+y_2\,cos\,\theta)\,\sigma_2^2\,e_2^2$
Note that
$e_1^2 = cos\,\theta\;e_1^1 + sin\,\theta\;e_2^1$
$e_2^2 = -sin\,\theta\;e_1^1 + cos\,\theta\;e_2^1$
So, we have:
$\Sigma_1\,v_1+A\,v_2 = x_1\,\sigma_1^1\,e_1^1+y_1\,\sigma_2^1\,e_2^1 + (x_2\,cos\,\theta+y_2\,sin\,\theta)\,\sigma_1^2\,(cos\,\theta\;e_1^1 + sin\,\theta\;e_2^1) + (-x_2\,sin\,\theta+y_2\,cos\,\theta)\,\sigma_2^2\,(-sin\,\theta\;e_1^1 + cos\,\theta\;e_2^1) = \\
\boxed{(x_1\,\sigma_1^1 + x_2\,cos^2\,\theta\;\sigma_1^2+y_2\,sin\,\theta\;cos\,\theta\;\sigma_1^2 + x_2\,sin^2\,\theta\;\sigma_2^2-y_2\,sin\,\theta\;cos\,\theta\;\sigma_2^2)\,e_1^1 + \\
(y_1\,\sigma_2^1 + x_2\,sin\,\theta\;cos\,\theta\;\sigma_1^2+y_2\,sin^2\,\theta\;\sigma_1^2 - x_2\,sin\,\theta\;cos\,\theta\;\sigma_2^2+y_2\,cos^2\,\theta\;\sigma_2^2)\,e_2^1}\;\;(1)$
And also:
$\boxed{\sigma_1^1\,(v_1+v_2) = (x_1\,\sigma_1^1 + x_2\,\sigma_1^1)\,e_1^1 + (y_1\,\sigma_1^1 + y_2\,\sigma_1^1)\,e_2^1}\;\;(2)$
So, if I did not commit any mistakes, we need to show that the norm of $(2)$ is always greater (or equal) the norm of $(1)$.
