Minimum value of this expression , was my inequality true or of any help? The problem which I encountered was a,b,c>0 and

$$f(a,b,c):= \left\lfloor\frac{a+b}{c}\right\rfloor+ \left\lfloor\frac{b+c}{a}\right\rfloor+ \left\lfloor\frac{c+a}{b}\right\rfloor$$
where $a$, $b$ and $c$ are positive. Find minimum of $f(a,b,c)$. For what values of $a$, $b$ and $c$ is the function minimum?

Is there a inequality like $\left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\geq \left\lfloor x+y\right\rfloor$ (and its proof) something which can be helpful here?
As such what I tried to get, expression is equivalent to $\newcommand{\lf}{\left\lfloor}\newcommand{\rf}{\right\rfloor}$
$$\lf\frac{a+b+c}{a}\rf + \lf\frac{a+b+c}{b}\rf + \lf\frac{a+b+c}{c}\rf-3$$
I don't know if that is correct. Can anyone help me?
 A: Since $\lfloor x\rfloor\gt x-1$, we have, by AM-GM inequality,
$$\begin{align}f(a,b,c)&=\left\lfloor\frac{a+b}{c}\right\rfloor+ \left\lfloor\frac{b+c}{a}\right\rfloor+ \left\lfloor\frac{c+a}{b}\right\rfloor
\\\\&\gt \frac{a+b}{c}-1+ \frac{b+c}{a}-1+ \frac{c+a}{b}-1
\\\\&=\bigg(\frac ba+\frac ab\bigg)+\bigg(\frac cb+\frac bc\bigg)+\bigg(\frac ac+\frac ca\bigg)-3
\\\\&\ge 2\sqrt{\frac ba\cdot\frac ab}+2\sqrt{\frac cb\cdot\frac bc}+2\sqrt{\frac ac\cdot\frac ca}-3
\\\\&=3\end{align}$$
from which we have
$$f(a,b,c)\ge 4$$
whose equality is attained when
$$(a,b,c)=(0.9,1,1)$$
Hence, the minimum of $f(a,b,c)$ is $\color{red}4$.

Added :
Under the condition that $0\lt a\le b\le c$, $f(a,b,c)=4$ holds if and only if $(a,b,c)$ satisfies
$$1\lt \frac ba\lt \frac 32\qquad\text{and}\qquad c\lt \min(3a-b,2b-a)$$
Proof :
We have
$$\left\lfloor\frac{b+c}{a}\right\rfloor\ge \left\lfloor\frac{c+a}{b}\right\rfloor\ge\left\lfloor\frac{a+b}{c}\right\rfloor\ge 0,\qquad  \left\lfloor\frac{b+c}{a}\right\rfloor\ge 2,\qquad \left\lfloor\frac{c+a}{b}\right\rfloor\ge 1$$
Suppose that $\bigg(\left\lfloor\dfrac{a+b}{c}\right\rfloor,\left\lfloor\dfrac{b+c}{a}\right\rfloor,\left\lfloor\dfrac{c+a}{b}\right\rfloor\bigg)=(0,3,1)$. Then, $$\begin{align}&\dfrac{a+b}{c}\lt 1,\qquad \dfrac{b+c}{a}\lt 4,\qquad \dfrac{c+a}{b}\lt 2
\\\\&\implies a+b\lt c,\qquad c\lt 4a-b,\qquad c\lt 2b-a
\\\\&\implies a+b\lt 4a-b,\qquad a+b\lt 2b-a
\\\\&\implies 2a\lt b\lt \frac 32a\end{align}$$
which is impossible.
Suppose that $\bigg(\left\lfloor\dfrac{a+b}{c}\right\rfloor,\left\lfloor\dfrac{b+c}{a}\right\rfloor,\left\lfloor\dfrac{c+a}{b}\right\rfloor\bigg)=(0,2,2)$. Then, $$\dfrac{a+b}{c}\lt 1,\qquad \dfrac{b+c}{a}\lt 3$$
$$\implies a+b\lt c\lt 3a-b\implies a+b\lt 3a-b\implies b\lt a$$
which is impossible.
So, we have to have $\bigg(\left\lfloor\dfrac{a+b}{c}\right\rfloor,\left\lfloor\dfrac{b+c}{a}\right\rfloor,\left\lfloor\dfrac{c+a}{b}\right\rfloor\bigg)=(1,2,1)$ which is equivalent to
$$\begin{align}&1\le\frac{a+b}{c}\lt 2,\qquad 2\le\frac{b+c}{a}\lt 3,\qquad 1\le\frac{c+a}{b}\lt 2
\\\\&\iff \frac{a+b}{2}\lt c\le a+b,\qquad 2a-b\le c\lt 3a-b,\qquad b-a\le c\lt 2b-a
\\\\&\iff 1\lt \frac ba\lt \frac 32\qquad\text{and}\qquad c\lt \min(3a-b,2b-a)\end{align}$$
