Let $a,b,c,x,y,z>0$. Under which condition of $a,b,c$ there will exists some $x,y,z>0$ such that \begin{equation} \begin{cases} x - \dfrac{a}{2} - c \left( 4x^2 + y^2 + z^2 \right) & > 0 \\ y - b - c \left( 4x^2 + y^2 + z^2 \right) & > 0 \\ z - c \left( 4x^2 + y^2 + z^2 \right) & > 0 \end{cases} \end{equation}
I know that $a,b,c$ should be somehow sufficiently small. But how small is it in a precise way?
My first observation is that we can multiply the first inequality by $2$ to consider instead the following system \begin{equation} \begin{cases} x' - a - 2 c \left( \left( x' \right) ^2 + y^2 + z^2 \right) & > 0 \\ y - b - c \left( \left( x' \right) ^2 + y^2 + z^2 \right) & > 0 \\ z - c \left( \left( x' \right) ^2 + y^2 + z^2 \right) & > 0 \end{cases} \end{equation} Then if we consider $U := \left( x' , y , z \right)$ where $x',y,z>0$ and $x'=2x$. We got something like \begin{equation} \left\lVert U \right\rVert ^{2} < \dfrac{1}{c} \min \left\lbrace \dfrac{1}{2} \left( x'-a \right) , y-b , z \right\rbrace . \end{equation} So the solution $U := \left( x' , y , z \right)$ should have the norm square less than some quatities which related to each components. Perhaps there should be some geometrical explaination for this or I simply just going in the wrong direction?