Upper bound on smallest singular value with subset condition Let $v_1,v_2,v_3\in \mathbb{R}^2$ (view as column vectors), if we have
$$\lambda_{\min}(v_iv_i^T+v_jv_j^T)\le 1,~\text{for all }i\not=j\in\{1,2,3\}.$$
Can we have an upper bound for
$$\lambda_{\min}(v_1v_1^T+v_2v_2^T+v_3v_3^T).$$
I guess it will be upper bounded by $3$ but have not idea how to prove it. Any idea will be appreciate.
 A: The upper bound is $\geq 3$. 
Take $v_1=[\sqrt{2},0]^T,v_2=Rot(v_1,2\pi/3),v_3=Rot(v_1,-2\pi/3)$. Then $\lambda_{\min}(v_1v_1^T+v_2v_2^T+v_3v_3^T)=3$.
EDIT. A partial converse for "the upper bound is indeed $3$". 
If $v_j=[\alpha_j,\beta_j]^T$, then we put $z_j=\alpha_j+i\beta_j$. The $3$ constraints can be written 
$\lambda_{min}(v_jv_j^T+v_kv_k^T)=1/2(|z_j^2|+|z_k^2|-|z_j^2+z_k^2|)\leq 1$.
The function to maximize can be written 
$g=\lambda_{min}(v_1v_1^T+v_2v_2^T+v_3v_3^T)=1/2(|z_1^2|+|z_2^2|+|z_3^2|-|z_1^2+z_2^2+z_3^2|)$.
We change the complex $z_j^2$ into the complex $u_j=a_j+ib_j$. Finally, the $3$ constraints are 
$\sqrt{a_j^2+b_j^2}+\sqrt{a_k^2+b_k^2}-\sqrt{(a_j+a_k)^2+(b_j+b_k)^2}\leq 2$
and the function to maximize is
$f(a_j,b_j)=2g=\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2}+\sqrt{a_3^2+b_3^2}-\sqrt{(a_1+a_2+a_3)^2+(b_1+b_2+b_3)^2}$.
Numerical experiments, using the Maple's software "optimization", seem to show that $sup(f)=6$ and is reached only by any rotation of the figure described in the first part of my post (line 2).
