[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 $(1)$.