This is with reference to example 2.21 in Boyd's book on Convex Optimization. I am attaching a screenshot below
Here, I do not understand how we get the conditions on $\mu$ and $\lambda$. To be precise, I do not understand how the implication (highlighted in gray) works in the example.
Could someone explain the reason behind this?
If I understand your question correctly, you are asking: why does $\lambda^{T}y \geq \mu$ for all $y \succ 0$ imply that $\mu \leq 0$ and $\lambda \succeq 0$? ($\lambda \neq 0$ is by hypothesis)
To begin, consider the left-hand side of $\lambda^{T}y \geq \mu$. We can make $\lambda^{T}y$ as close to zero as we like, by taking $y = (\frac{1}{n}, \cdots, \frac{1}{n})$ and letting $n \to \infty$. In the limit, the equality then becomes $0 \geq \mu$.
Next, and again considering $\lambda^{T}y \geq \mu$, if there was some entry of $\lambda$ that was strictly negative, say the first entry, then we could take $y = (n,\frac{1}{n}, \cdots, \frac{1}{n})$, so that $\lambda^{T}y \to -\infty$ as $n \to \infty$, and it is impossible for $\lambda^{T}y$ to be bounded below by a fixed $\mu$.