Let $\Lambda(x,y)$ be the relative totient function that counts the amount of numbers less than $x$, which are coprime to $y$.
After interpreting Euler's totient function, $\phi(y)$, as a result of combinatorics and modulo arithmetic, I've tried to interpret $\Lambda(x,y)$ as a result of combinatorics and modulo arithmetic too and reached the following form, I would like to complete:
$$\Lambda(x,y) \geq \frac{x}{y}\phi(y) + V$$
Obviously, if $x$ is a multiple of $y$, then $V=0$, because that's a well established fact. But considering $x$ as a non-multiple of $y$; I note that $\phi(y)$ has product form derivable from combinatorics and modulo arithmetic, and that any term in that product involving a prime divisor of both $x$ and $y$ cannot produce error. But any term in that product involving a prime divisor of just $y$ can produce error. So the question is how much, and consequently, what is $V$?
In the worst case, for any prime $p$ divisor of only $y$ and not $x$, the corresponding term in Euler's product would be errorless up to a value, let's label $c_{p}$, where $c_{p}$ is the largest multiple of $p$ less than $x$. So the largest error that the term in Euler's product can make in this relative totient function formulation, is to predict too many totatives of $p$ between $c_{p}$ and $x$, so there's an argument that the upper bound to that amount would be $p – 1$, the largest possible remainder when dividing by $p$.
After considering all of the primes divisors of $y$ which are not divisors of $x$; the prime $q$ which is the largest of those primes, can feasibly cause the largest error, because $c_{q}$ could be the maximum distance away from $x$. Assuming the worst case, that $c_{q} = x – q + 1$, then we can argue that the upper bound of the amount of extra totatives predicted by Euler's product is $q-1$, and any error due to other primes $p$, which are only divisors of $y$, is subsumed by the error caused by $q$, because $c_{p} > c_{q}$ and all $p-1$ at most incorrect totatives, are already included in this formulation as incorrect totatives of $q$.
So $V\geq -(q-1)$ to ammend the error derived above, and $$\Lambda(x,y) \geq \frac{x}{y}\phi(y) – q$$ where $q$ is the largest prime divisor of $y$ but not $x$.
How sound is this analysis?
EDIT
This is a more formal version.
Lemma $$\Lambda(x,y) > \frac{x}{y} \phi(y) – z$$ where $z$ is the largest prime that divides $y$, but doesn't divide $x$, $y$ is larger than $x$, and $x$ is larger than all prime divisors of $y$
Let $D_y := \{d : (d,y) \neq 1 \}$ and $P_y := \{p : p \in D_y \cap \mathbb{P} \}$ and $Q_{(x,y)} := \{q : q \in P_y$ and $q \mid x \}$ and $R_{(x,y)} := \{r : r \in P_y$ and $r \nmid x \}$.
I'm going to use some made up notation in the next part because I have no idea how to write this succinctly otherwise. Unfortunately, the left subscripts may shift a little, apologies in advance.
Let ${}_a\mid \Sigma p \mathbb{Z} \mid _b$ be the result of counting all elements in $p\mathbb{Z}$ bounded in the set $[a,b]$, for each $p \in P_y$. (So multiples may be counted more than one time).
Let ${}_a\mid D_y \mid _b$ be the amount of elements in the set $D_y$ that are bounded to the set $[a,b]$
With this style of notation, hopefully the rest expresses as I mean.
By exhaustion with duplication ${}_0\mid D_y \mid _y \leq {}_0\mid \Sigma p \mathbb{Z} \mid _y$.
By classifying primes then ${}_0\mid \Sigma p \mathbb{Z} \mid _y = {}_0\mid \Sigma q \mathbb{Z} \mid _y + {}_0\mid \Sigma r \mathbb{Z} \mid _y$ for $q \in Q_{(x,y)}$ and $r \in R_{(x,y)}$
By partitioning ${}_0\mid \Sigma p \mathbb{Z} \mid _y \leq {}_0\mid \Sigma q \mathbb{Z} \mid _x + {}_x\mid \Sigma q \mathbb{Z} \mid _y + {}_0\mid \Sigma r \mathbb{Z} \mid _x + {}_x\mid \Sigma r \mathbb{Z} \mid _y $ for $q \in Q_{(x,y)}$ and $r \in R_{(x,y)}$.
Then ${}_0\mid D_y \mid _y \leq {}_0\mid \Sigma q \mathbb{Z} \mid _x + {}_x\mid \Sigma q \mathbb{Z} \mid _y + {}_0\mid \Sigma r \mathbb{Z} \mid _x + {}_x\mid \Sigma r \mathbb{Z} \mid _y $
It follows by partitioning that ${}_0\mid D_y \mid _x \leq {}_0\mid \Sigma q \mathbb{Z} \mid _x + {}_0\mid \Sigma r \mathbb{Z} \mid _x$.
By definition $\Lambda(x,y) = x – {}_0\mid D_y \mid _x$
So $\Lambda(x,y) \geq x – {}_0\mid \Sigma q \mathbb{Z} \mid _x – {}_0\mid \Sigma r \mathbb{Z} \mid _x$.
Let ${}_a\mid \{ \Sigma p \mathbb{Z} \} \mid _b$ be the result of counting all elements in $p\mathbb{Z}$ bounded in the set $[a,b]$, for each $p \in P_y$, but not counting duplicates.
Then it still must be true that $\Lambda(x,y) \geq x – {}_0\mid \{ \Sigma q \mathbb{Z} \}\mid _x – {}_0\mid \{ \Sigma r \mathbb{Z} \} \mid _x$ because some non totatives of $y$ may be multiples of elements from both $Q_{(x,y)} $ and $R_{(x,y)} $
Note as well that $\phi(y) = y – {}_0\mid D_y \mid _y$.
Then $\phi(y) = y – {}_0\mid \{ \Sigma q \mathbb{Z} \} \mid _y – \mid Z_{(x,y)} \mid $ where $Z_{(x,y)}$ is a subset of $\{ \Sigma r \mathbb{Z} \} $ that shares no elements with $\{ \Sigma q \mathbb{Z} \} $.
So $ \frac{x}{y}\phi(y) = x – \frac{x}{y}{}_0\mid \{\Sigma q \mathbb{Z} \} \mid _y – \frac{x}{y} \mid Z_{(x,y)} \mid$
And $ \frac{x}{y}\phi(y) = x – {}_0\mid \{\Sigma q \mathbb{Z} \} \mid _x – \frac{x}{y} \mid Z_{(x,y)} \mid$
The proof attempt stops here, so I think I answered my own question.
Best Answer
About the bound you obtained $$ \Lambda(x,y) \geq \frac{x}{y}\phi(y) - q $$ where $q$ is the largest prime divisor of $y$ but not $x$, this lower bound becomes useless if $q=y$ is prime larger than $x$. To see this, the lower bound becomes negative. $$ \frac xy\phi(y) -q \leq x-q <0. $$
To amend this situation, we must have a good error term between $\Lambda(x,y)$ and $\frac xy \phi(y)$. A standard method is applying Mobius function $\mu(n)$.
$$ \begin{align} \Lambda(x,y)&=\sum_{\substack{{n\leq x}\\{(n,y)=1}}} 1=\sum_{n\leq x} \sum_{\substack{{d|n}\\{d|y}}}\mu(d)\\ &=\sum_{d|y} \mu(d)\sum_{k\leq \frac xd} 1 \ \ (\text{by }n=dk)\\ &=\sum_{d|y} \mu(d)\left(\frac xd - \left\{ \frac xd \right\}\right)\\ &=\sum_{d|y} x \frac{\mu(d)}d- \sum_{d|y}\mu(d)\left\{ \frac xd \right\} \ \ (\{a\}\text{ is the fractional part of }a, \ \lfloor a \rfloor = a - \{a\}. )\\ &=\frac xy \phi(y)-\sum_{d|y}\mu(d)\left\{ \frac xd \right\}.\end{align} $$ The error term here is $$ E(x,y)=-\sum_{d|y}\mu(d)\left\{ \frac xd \right\}. $$ This error term has an estimate $$ |E(x,y)|\leq \sum_{d|y}|\mu(d)|=2^{\omega(y)} $$ where $\omega(y)$ is the number of distinct prime divisors of $y$. It is also used sometimes that $2^{\omega(y)}\leq \tau(y)$ with the number-of-divisor function $\tau(y)$.
Hence, we have the lower bound of $\Lambda(x,y)$ given as $$ \Lambda(x,y)\geq \frac xy \phi(y) - 2^{\omega(y)} $$ or $$ \Lambda(x,y)\geq \frac xy \phi(y)-\tau(y). $$ An advantage of these bounds are that $\tau(y)$ is small compared to $y$. In fact, $$ \tau(y)\leq y^{\frac{\log 2}{\log\log y} + O\left(\frac1{(\log\log y)^2}\right)}. $$