Given negatives $x$ and $y$ such that $\left(\dfrac{x}{2} – 1\right)^2 + \left(\dfrac{y}{2} – 1\right)^2 \le \dfrac{125}{2}$. Calculate the maximum value of $$\large \lfloor x \lfloor x \rfloor \rfloor + \lfloor y \lfloor y \rfloor \rfloor$$
We could solve for the maximum value of $x^2 + y^2$.
We have that $$\left(\frac{x}{2} – 1\right)^2 + \left(\frac{y}{2} – 1\right)^2 \le \frac{125}{2} \iff \frac{x^2 + y^2}{4} – (x + y) + 2 \le \frac{125}{2}$$
$$\iff x^2 + y^2 \le 2[121 – 2(x + y)]$$
Moreover, $$\frac{(x + y)^2}{8} – (x + y) – \frac{121}{2} \le 0 \implies x + y \in (4 – 10\sqrt5, 0)$$ since $x, y < 0$.
$$\implies x^2 + y^2 \le 2[121 – 2(4 – 10\sqrt5)] = 2(113 + 20\sqrt5)$$
But I'm uncertain about the case for $x \lfloor x \rfloor + y \lfloor y \rfloor$ or even $\lfloor x \lfloor x \rfloor \rfloor + \lfloor y \lfloor y \rfloor \rfloor$.
Best Answer
Let us use the following lemma :
Lemma : To find the maximum value of $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor$, we only need to consider $(x,y)$ such that $(x-2)^2+(y-2)^2=250$.
The proof of the lemma is written at the end of the answer.
From the lemma, we have $$y=2-\sqrt{250-(x-2)^2}$$ which is decreasing for $x\lt 0$.
We may suppose that $x\ge y$. Now, solving the system $$250-(x-2)^2\ge 0\qquad\text{and}\qquad x\lt 0\qquad\text{and}\qquad y\lt 0\qquad\text{and}\qquad x\ge y$$ gives $$(-9.18\approx)\ 2 - 5 \sqrt 5\le x\lt 0$$
If $-1\le x\lt 0$, then we get$$y\gt 2-\sqrt{250-(0-2)^2}=2-\sqrt{246}\gt 2-\sqrt{246.49}=2-15.7=-13.7$$Since $\lfloor x\rfloor=-1$, we get $x\lfloor x\rfloor\le 1$ implying $\lfloor x\lfloor x\rfloor\rfloor\le 1$. Since $\lfloor y\rfloor\ge -14$, we get $y\lfloor y\rfloor\lt (-14)\times (-13.7)=191.8$ implying $\lfloor y\lfloor y\rfloor\rfloor\le 191$. So, $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 1+191=192$
If $-2\lt x\lt -1$, then $-13.53\lt y$, so $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 3+189=192$
If $-\frac 73\lt x\le -2$, then $-13.3\lt y$, so $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 6+186=192$
If $-3\lt x\le -\frac 73$, then $-13.21\lt y$, so $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 8+184=192$
If $-4\lt x\le -3$, then $-13\le y$, so $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 15+169=184$
If $-5\le x\le -4$, then $-12.7\lt y$, so $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 25+165= 190$
If $-\frac{17}{3}\le x\lt -5$, then $-12.2\lt y$, so $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 34+158= 192$
If $-7\lt x\lt -\frac{17}{3}$, then $-11.9\lt y$, so $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 48+142=190$
If $-8\le x\le -7$, then $-11\le y$, so $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 64+121=185$
If $-\frac{79}{9}\le x\lt -8$, then $-10.3\lt y$, so $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 79+113=192$
If $-9\le x\lt -\frac{79}{9}$, then $-9.6\lt y$, so $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 81+96=177$
If $2-5\sqrt 5\le x\lt -9$, then $-9.4\lt y$, so $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 91+94=185$
It follows from these that $$\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 192$$ whose equality is attained when $$(x,y)=\bigg(-\frac{201}{100},\frac{200-9\sqrt{28879}}{100}\bigg)$$
Therefore, the maximum value of $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor$ is $\color{red}{192}$.
Finally, let us prove the lemma.
Lemma : To find the maximum value of $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor$, we only need to consider $(x,y)$ such that $(x-2)^2+(y-2)^2=250$.
Proof for lemma :
Let $m$ be a negative integer. Also, let $\alpha$ be a real number such that $0\le \alpha\lt 1$.
To prove the lemma, it is sufficient to prove the followings :
(1) For any fixed $m$, $f(\alpha):=\lfloor (m+\alpha)\lfloor (m+\alpha)\rfloor\rfloor$ is decreasing.
(2) For any $(m,\alpha)$, $\lfloor (m+\alpha)\lfloor m+\alpha\rfloor\rfloor\ge \lfloor (m+1)\lfloor m+1\rfloor\rfloor$
Proof for (1) : $$f(\alpha)=\lfloor (m+\alpha)\lfloor (m+\alpha)\rfloor\rfloor=m^2+\lfloor m\alpha\rfloor$$is decreasing.
Proof for (2) :
$$\begin{align}&\lfloor (m+\alpha)\lfloor m+\alpha\rfloor\rfloor- \lfloor (m+1)\lfloor m+1\rfloor\rfloor \\\\&=m^2+\lfloor m\alpha\rfloor-(m+1)^2 \\\\&=-2m-1+\lfloor m\alpha\rfloor \\\\&\ge -2m-1+m \\\\&\ge 0\qquad\square\end{align}$$