Showing that $(x_2 - x_1)^2 + (y_2 - y_1)^2 \leq 8$, where $x^2 + y^2 \leq 1$

Question

Let $D^2$ be the unit disk in $\mathbb{R}^2$ and $(x_1, y_1), (x_2, y_2) \in D^2$. I can't show that $$(x_2 - x_1)^2 + (y_2 - y_1)^2 \leq 8.$$ Can someone give me a hint?

Answer 1

Here's a more circuitous, but more general, argument using the triangle inequality several times and the fact that if a point lies inside the unit disk, all of its coordinates must have magnitude no greater than $1$: \begin{align*} (x_2 - x_1)^2 + (y_2-y_1)^2 &=\big| (x_2 - x_1)^2 + (y_2-y_1)^2\big| \\ &= \big|x_2^2 - 2x_1x_2 + x_1^2 + y_2^2 - 2y_1y_2 + y_1^2\big| \\ &= \big|(x_1^2 + y_1^2) + (x_2^2 + y_2^2) - 2(x_1x_2 + y_1y_2)\big| \\ &\leq \big|x_1^2\big| + \big|y_1^2\big| + \big|x_2^2\big| + \big|y_2^2\big| + 2\bigg(\big|x_1x_2\big| + \big|y_1y_2\big|\bigg)\\ &\leq 1 + 1 + 1 + 1 + 2(|x_1||x_2|+|y_1||y_2|)\\ &\leq 4 + 2(1 + 1)\\ &= 8 \end{align*}

(Note that this argument actually holds for any point in the unit square, not just the unit disk.)

Answer 2

If we apply the euclidean norm, as we have that the maximum distance between the pair of coordinates $x_1,x_2$ and $y_1,y_2$ is 2, meaning that they are at the opposite extremes of the disc, we can then show:

$$ \lVert (x_1 - x_2)^2\rVert + \lVert (y_1 - y_2)^2 \rVert \leq 2^2 + 2^2 = 8$$

Answer 3

The inequality can actually be tighter... The geometric argument is straightforward: $$ (x_2-x_1)^2+(y_2-y_1)^2 $$ is the squared distance between the two points. As these lay inside the unit disk, their distance is as most 2, therefore $$ (x_2-x_1)^2+(y_2-y_1)^2\leq 4. $$

Answer 4

By C-S $$(x_1-x_2)^2+(y_1-y_2)^2=x_1^2+y_1^2+x_2^2+y_2^2-2(x_1x_2+y_1y_2)\leq$$ $$\leq2+2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}\leq4<8$$