Let $f$ be any multiplicative function with $|f(n)| \le 1$ and such that $\sum_{d|n} f(d)$ is non-negative for all $n$. It is easy to check that $\mu$ and $\lambda$ satisfy this constraint. Then $$ 0\le \sum_{n\le x} \sum_{d|n} f(d) = \sum_{d\le x} f(d) \lfloor \frac{x}{d} \rfloor \le \sum_{d\le x} \Big( x\frac{f(d)}{d} + 1\Big), $$ so that $$ \sum_{d\le x} \frac{f(d)}{d} \ge - \frac{\lfloor x\rfloor}{x} \ge -1. $$
The upper bound for $\mu$ and $\lambda$ follows similarly, here making use of $\sum_{d|n} \mu(d) =1$ if $n=1$ and $0$ otherwise, and $\sum_{d|n} \lambda(d) = 1$ if $n$ is a square and zero otherwise. So for $\lambda$ we obtain $$ \sum_{n\le x} \frac{\lambda(n)}{n} \le \frac{x+\sqrt{x}}{x}. $$
For a more thorough discussion of such partial sums (especially with regard to the general lower bound), see this paper of Granville and Soundararajan.