I have a problem with a system of non-linear inequalities:
\begin{cases} (x-\frac{y}{B})(y-x-1) \ge \frac{(y-1)x}{2} \\ y \ge 2 \end{cases}
and I would like to find the exact value of the maximum of the constant $B \ge 1$ for which it doesn't have solutions in $x,y$. Using Wolfram Alpha I know that $\max B \gt 11.6$.
Indeed the answer is $6+4\sqrt{2}\approx 11.65685$. To see this, Let $\mathcal{B}$ be the set of real numbers $B$ such that for all $y\ge2$ and all $x\in\mathbb{R}$ we have $$\left(x-\frac{y}{B}\right)(y-x-1)<\frac{(y-1)x}{2}.\tag{1}$$ $B\in\mathcal{B}~$ if and only if, for all $y\ge2$, we have $$\forall\, x\in\mathbb{R},\quad\left(x-\frac{y}{B}\right)(y-x-1)<\frac{(y-1)x}{2},$$ or equivalently $$\forall\, x\in\mathbb{R},\quad x^2-\left(\frac{y-1}{2}+\frac{y}{B}\right)x+\frac{y(y-1)}{B}>0.$$ That is, the discriminant of this second degree equation is negative: $$P(y)\triangleq\left(\frac{y-1}{2}+\frac{y}{B}\right)^2-\frac{4y(y-1)}{B}<0.$$ Note that $\deg P(y)\le 2$, with the coefficient of $y^2$ being $\left(\frac{1}{2}+\frac{1}{B}\right)^2-\frac{4}{B}$. Moreover, $P(1)=1/B^2>0$. Now $P(y)<0$ for all $y\ge2$, if and only if, $\left(\frac{1}{2}+\frac{1}{B}\right)^2-\frac{4}{B}\le0$ and $P(2)<0$. The first condition is equivalent to $B\in[6-4\sqrt{2},6+4\sqrt{2}]$ and the second one is equivalent to $\frac{B}{2}\in(6-4\sqrt{2},6+4\sqrt{2})$. Hence $$\mathcal{B}=\left(12-8\sqrt{2},6+4\sqrt{2}\,\right].$$ In particular, $\max(\mathcal{B})=6+4\sqrt2$ as announced.