I have the following system of inequalities:($x,y,z\in\mathbb{R}$)
$ \begin{cases} \Phi\geq\frac{1}{2}(x^2-y^2-z^2)+x+y+z-\frac{3}{8},\\ \Phi\leq \frac{1}{2}(x^2-y^2-z^2)+\frac{1}{2}x+y+z-\frac{9}{8}\\ \Phi\leq \frac{1}{2}(x^2-y^2-z^2)+\frac{1}{8}\\ y\geq\frac{1}{2}, z\geq 1 \end{cases} $
The task is find the range of $\Phi$ such that this system of inequalities in $x,y,z$ has solutions in $\mathbb{R}$.
I am wondering what is the rigorous way to do such kind of problem. I realized this might be some kind of high school problem. Thanks in advance.
There is no upper limit for $\Phi$, as $x$ can be chosen arbitrarily high to fulfil the second and the third inequality.
The first inequality can be rewritten to: $$ 2\Phi\geq(x^2+2x+1)-(y^2-2y+1)-(z^2-2z+1)+\frac{1}{4} $$ Inserting the inequalities for $y$ and $z$: $$ 2\Phi\geq(x+1)^2 $$
Therefore, $\Phi$ must be larger than or equal to zero.