Smooth approximation of absolute value inequalities

Is there an analytic approximation to the inequality:

$$\sum_{i=1}^{n} |x_i| \leq \delta ? $$

I would like to replace the above inequality with a smooth inequality that is "valid" in the sense that if the approximate smooth inequality is satisfied then the original inequality will also be satisfied. It would also be great if the approximating inequality is tight. I have an intuitive idea of "tightness" but don't know how to formalize it for $n > 1$.

Solution 1:

For every positive $u$, the functions $f_u$ and $g_u$ defined by $$ f_u(x)=\frac{x^2}{\sqrt{x^2+u^2}},\qquad g_u(x)=\sqrt{x^2+u^2}, $$ are smooth and such that, for every $x$, $$f_u(x)\leqslant|x|\leqslant g_u(x).$$Furthermore, the error one makes when replacing $|x|$ by $f_u(x)$ or by $g_u(x)$ is at most $$ g_u(x)-f_u(x)=\frac{u^2}{\sqrt{x^2+u^2}}\leqslant u, $$ uniformly on $x$.

Hence, for small (quantifiable) positive values of $u$, the functions $$\sum\limits_{i=1}^nf_u(x_i)\qquad\text{and/or}\qquad\sum\limits_{i=1}^ng_u(x_i)$$ are smooth and accurate approximations of $$\sum\limits_{i=1}^n|x_i|$$ Every $u\leqslant\varepsilon/n$ yields a uniform error which is at most $\varepsilon$, thus all this yields, for example, for every $x$, $$\left|\sum\limits_{i=1}^n|x_i|-f(x)\right|\leqslant\varepsilon$$ where $$f(x)=\sum\limits_{i=1}^n\frac{nx_i^2}{\sqrt{n^2x_i^2+\varepsilon^2}}$$