Is $\tau(x,y)\leq\sqrt{n}d(x,y)$ where $d$ is the Euclidean metric, and $\tau$ the taxicab metric?

Suppose $d$ is the Euclidean metric on $\mathbb{R}^n$, $d(x,y)=\sqrt{\sum_{i=1}^n(x_i-y_i)^2}$, and $\tau(x,y)=\sum_{i=1}^n|x_i-y_i|$ is the taxicab metric on $\mathbb{R}^n$.

When showing this two metrics are equivalent, it is not hard to bound $\tau$ by $$\tau(x,y)\leq n d(x,y).$$

Is it possible to tighten the bound to $\tau(x,y)\leq\sqrt{n}d(x,y)$?

No, it's not possible: $\tau( (0.4,0.4), (0,0) ) = 0.8 > 0.46 > \sqrt{2} (0.4^2+0.4^2) = \sqrt{2}d( (0.4,0.4), (0,0) )$
edit: As I somehow missed to take the square root, this counterexample is wrong. See below for a correct answer that it is in fact possible to work with $\sqrt{n}$.
I think the counterexample forgot to take the square root of $d$. Instead, this follows from the Cauchy-Schwarz inequality.
If $x,y$ are in $\mathbb{R}^n$, $$\tau(x,y)=\sum_{i=1}^n|x_i-y_i|=\left|\sum_{i=1}^n|x_i-y_i|\right|\leq\sqrt{\sum_{i=1}^n (x_i-y_i)^2}\sqrt{\sum_{i=1}^n 1^2}=\sqrt{n}d(x,y).$$