Understanding an inequality in $\mathbb{R}^{n}$ Let $Q = Q(x_0,r)$ be a cube with sides parallel to the coordinate axes, diameter $r>0$ and center $x_0\in \mathbb{R}^{n}$. Let $Q^{\prime} = Q(y_0,r)$. Let $B = B(x_0,r)$ be a ball in $\mathbb{R}^{n}$ of radius $r$ centered at $x_0$. Let $B^{\prime} = B(y_0,r)$. By taking $x \in Q$, $y \in Q^{\prime}$ or $x \in B$, $y \in B^{\prime}$ we obtain that
$$
\left|\frac{x-x_0}{r}\right|+\left|\frac{y-y_0}{r}\right|\le \sqrt{n}.
$$ 
I can't understand this inequality? Could you please explain me?

A cube with diameter $r$ will be completely contained in a ball of radius $r$. So by showing the statement for the ball, the statement for the cube follows. For the ball, $\left\lvert\frac{x-x_0}r\right\rvert\le1$. This bound is tight, it's essentially the definition of a ball. So
$$\left\lvert\frac{x-x_0}r\right\rvert+\left\lvert\frac{y-y_0}r\right\rvert\le2=\sqrt4$$
which makes the claim wrong for $n<4$.
So back to the cube. diameter $r$ means it fits in a ball of radius $\frac r2$. In other words, $\left\lvert\frac{x-x_0}r\right\rvert\le\frac12$ and thus
$$\left\lvert\frac{x-x_0}r\right\rvert+\left\lvert\frac{y-y_0}r\right\rvert\le1=\sqrt1$$
Much better. Could it be that for the ball you need the diameter as well, not the radius?
