I know that for a Noetherian local ring $(R,\frak{m})$ and $a\in \frak{m}$ one can prove: $$\dim R/(a) \geq \dim R -1$$ by using parameters: let $d=\dim R/(a)$ and $x_1,\ldots,x_d\in R$ be a system of parameters for $\frak{m}$ in $R/(a)$. Thus the ideal $(a,x_1,\ldots,x_d)$ is $\frak{m}$-primary so by Krull Dimension Theorem we have $\dim R =\text{ht}\ \mathfrak{m}\leq 1+d$.
However, the textbook I'm reading Algebraic Geometry and Commutative Algebra, by Siegfried Bosch, Exercise 3, page 82, suggests that I should proceed otherwise: take a chain of primes in $R$: $$\mathfrak{p}_0\subsetneq\cdots\subsetneq \mathfrak{p}_n$$ such that $a\in \mathfrak{p}_n$, then we should be able to find a chain: $$\mathfrak{p}'_1\subsetneq\cdots\subsetneq \mathfrak{p}'_n$$ such that $a\in\mathfrak{p}'_1$ and $\mathfrak{p}'_n=\mathfrak{p}_n$. The exercise suggests induction.
I can't seem to prove it. Any ideas?
Edit: After some time, I realized I did not fully understand the question asked, but I did manage to find an answer that I'll post below, for completeness sake. The explicit question is the following:
Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak{m}$. Show for any $a_1,\ldots,a_r\in R$ that $\dim R/(a_1,\ldots,a_r)\geq\dim R-r$.
Hint: Assume $r=1$. Show for any chain of prime ideals $\mathfrak{p}_0\subsetneq\mathfrak{p}_1\subsetneq\ldots\subsetneq\mathfrak{p}_n$ where $a_1\in\mathfrak{p}_n$, that there is a chain of prime ideals $\mathfrak{p}_1'\subsetneq\ldots\subsetneq\mathfrak{p}_n'$ satisfying $a_1\in\mathfrak{p}_1'$ and $\mathfrak{p}_n'=\mathfrak{p}_n$; use induction on $n$.
Step 1: We first observe that the hint is enough to prove what is asked of us. Indeed for two elements $a,b\in R$ we have $$R/(a,b)=\frac{R/(a)}{b\cdot R/(a)}$$ so a quick induction argument would finish the job if we manage to prove the $r=1$ case. Let us thus prove the hint by induction.
Step 2: We see that the cases $n=0$ and $n=1$ are immediate. We will need to prove the case $n=2$ for the arbitrary induction step. So if $a_1\in\mathfrak{p}_2$ there is a minimal associated prime $\mathfrak{p}'_1\in\text{Ass}'((a_1))$ such that $\mathfrak{p}'_1\subset\mathfrak{p}_2$. But, by Krull's principal ideal theorem, we have $\text{ht}(\mathfrak{p}'_1)\leq 1$, but by hypothesis $\text{ht}(\mathfrak{p}_2)\geq2$ so the inclusion is strict.
Step 3: Take $n\in\mathbb N,\:n\geq2$ and assume our statement is true for any value below $n$; we wish to show that it is also true for $n+1$. Since $n\geq 2$ we can fix some $m\in\mathbb N$ such that $0<m<n$.
Case 1: $a_1\in\mathfrak{p}_m$. We then apply the induction hypothesis to the chain $\mathfrak{p}_0\subsetneq\ldots\subsetneq\mathfrak{p}_m$ and find a chain $\mathfrak{p}'_1\subsetneq\ldots\subsetneq\mathfrak{p}'_m=\mathfrak{p}_m$ such that $a_1\in\mathfrak{p}'_1$. We then simply concatenate this last one with the rest of the starting chain $\mathfrak{p}_{m+1}\subsetneq\ldots\subsetneq\mathfrak{p}_n$. to get the desired one.
Case 2: $a_1\notin\mathfrak{p}_m$. We consider a chain of ideals in $R/\mathfrak{p}_m$: $$(0)=\mathfrak{p}_m/\mathfrak{p}_m\subsetneq\mathfrak{p}_{m+1}/\mathfrak{p}_m\subsetneq\ldots\subsetneq\mathfrak{p}_n/\mathfrak{p}_m\subsetneq R/\mathfrak{p}_m.$$
The cahin is of length $n+1-m$ so we can apply the induction hypothesis and find a chain of primes $$\mathfrak{p}'_{m+1}/\mathfrak{p}_m\subsetneq\ldots\subsetneq\mathfrak{p}'_n/\mathfrak{p}_m=\mathfrak{p}_n/\mathfrak{p}_m,$$
(with $\mathfrak{p}_n\subset\mathfrak{p}'_{m+1}\subsetneq\ldots\subsetneq\mathfrak{p}'_n$) such that $a\mod\mathfrak{p}_m\in\mathfrak{p}'_{m+1}/\mathfrak{p}_n$, so we have $a_1\in\mathfrak{p}'_{m+1}$. Since $a_1\in\mathfrak{p}'_{m+1}\setminus\mathfrak{p}_n$ we have a strictly increasing chain of prime ideals $$\mathfrak{p}_0\subsetneq\ldots\subsetneq\mathfrak{p}_m\subsetneq\mathfrak{p}'_{m+1}$$
of length $m+1<n+1$, so we can apply the induction hypothesis and obtain a chain of ideals $$\mathfrak{p}'_1\subsetneq\ldots\subsetneq\mathfrak{p}'_m\subsetneq\mathfrak{p}'_{m+1}$$
with $a_1\in\mathfrak{p}'_1$, that we concatenate with the one found just above. This ends the proof of the hint.