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.

  • $\begingroup$ Have you found any non-rigorous way to do this? $\endgroup$ – Trebor Sep 17 '18 at 5:34
  • $\begingroup$ say, using mathematica to test some values for $\Phi$, then I guessed a range, say $\Phi\geq 0$, then, I prove within this range, the system of inequalities has solutions. But this only gave me some sufficient condition, I need a little bit more, say what is the precise range for $\Phi$. The problem I asked here is some kind of toy case, the real question has many variables. $\endgroup$ – user41650 Sep 17 '18 at 5:38

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.

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