For positive integer numbers, so $\min\{\frac{c}{a},\frac{c}{b}\}+\lfloor\frac{c}{a}\rfloor\lfloor\frac{c}{b}\rfloor\geqq c\lfloor\frac{c}{ab}\rfloor$ 
Given three positive integer numbers $a, b, c$. Prove that
  $$\min\left \{ \frac{c}{a}, \frac{c}{b} \right \}+ \left \lfloor \frac{c}{a} \right \rfloor\left \lfloor \frac{c}{b} \right \rfloor\geqq c\left \lfloor \frac{c}{ab} \right \rfloor$$

I think midway calculations are not easy formulas, I can't find what kind of formula transformation to solve! So I need to the help! Thanks for all the nice comments and interests from MathematicsSE-ees!
 A: The inequality requested to be proven is
$$\min\left( \frac{c}{a}, \frac{c}{b} \right) + \left\lfloor \frac{c}{a} \right\rfloor \left\lfloor \frac{c}{b} \right\rfloor \geqq c\left\lfloor \frac{c}{ab} \right\rfloor \tag{1}\label{eq1}$$
Since it's symmetric in $a$ and $b$, WLOG let $b \ge a$, so
$$\min\left(\frac{c}{a},\frac{c}{b}\right) = \frac{c}{b} \tag{2}\label{eq2}$$
Let
$$c = kab + d, \; k \ge 0, \; 0 \le d \lt ab \tag{3}\label{eq3}$$
The RHS of \eqref{eq1} becomes
$$c\left \lfloor \frac{c}{ab} \right \rfloor = (kab + d)(k) = abk^2 + kd \tag{4}\label{eq4}$$
Consider the case of $c \lt b$. This means $\left\lfloor \frac{c}{a} \right\rfloor \left\lfloor \frac{c}{b} \right\rfloor = 0$ and, since $a \ge 1$, also $c\left\lfloor \frac{c}{ab} \right\rfloor = 0$. Since $\frac{c}{b} \gt 0$, this means \eqref{eq1} holds with a strict inequality.
Otherwise, with $c \ge b$ (so $\frac{c}{a}$ and $\frac{c}{b}$ are each $\ge 1$, allowing the inequality below to subtract $1$ from each value and still have them be non-negative numbers), using \eqref{eq2}, the LHS of \eqref{eq1} becomes
\begin{align}
\frac{c}{b} + \left\lfloor \frac{c}{a} \right\rfloor \left\lfloor \frac{c}{b} \right\rfloor & = ka + \frac{d}{b} + \left\lfloor kb + \frac{d}{a} \right\rfloor \left\lfloor ka + \frac{d}{b} \right\rfloor \\
& \gt ka + \frac{d}{b} + \left(kb + \frac{d}{a} - 1\right)\left(ka + \frac{d}{b} - 1\right) \\
& = ka + \frac{d}{b} + abk^2 + kd - kb + kd + \frac{d^2}{ab} - \frac{d}{a} - ka - \frac{d}{b} + 1 \\
& = abk^2 + kd + \left(-kb + kd + \frac{d^2}{ab} - \frac{d}{a} + 1\right) \\
& = abk^2 + kd + \left(k(d-b) + \frac{d}{a}\left(\frac{d}{b} - 1\right) + 1\right) \tag{5}\label{eq5}
\end{align}
If $d \ge b$, then $k(d-b) + \frac{d}{a}\left(\frac{d}{b} - 1\right) + 1 \ge 1$, so the RHS of \eqref{eq5} is greater than that of \eqref{eq4}, meaning \eqref{eq1} holds with the strict inequality.
Next, consider the case where $d \lt b$. Let
$$d = ma + n, \; m \ge 0, \; 0 \le n \lt a \tag{6}\label{eq6}$$
Now, starting from the RHS of the first line of \eqref{eq5}, this gives
\begin{align}
ka + \frac{d}{b} + \left\lfloor kb + \frac{d}{a} \right\rfloor \left\lfloor ka + \frac{d}{b} \right\rfloor & = ka + \frac{d}{b} + (kb + m)(ka) \\
& \ge ka + (kb + m)(ka) \\
& = ka + abk^2 + mka \\
& = abk^2 + k(ma + a) \\
& \ge abk^2 + k(ma + n) \\
& = abk^2 + kd \tag{7}\label{eq7}
\end{align}
Once again, this shows that \eqref{eq1} holds. Note the only possibility for the first inequality in \eqref{eq7} to be an equals is if $d = 0$ since it's ignoring the $\frac{d}{b}$ term. Also, the only possibility of the second equality to be an equals is if $k = 0$ since $a \gt n$. Thus, both can be equals only if $k = d = 0$, but then this gives $c = 0$ in \eqref{eq3}, but the problem says $c \gt 0$. Thus, once again, it's a strict inequality in \eqref{eq1}.
As all possible cases have been considered, this shows \eqref{eq1} is always true, with it actually always having the LHS $\gt$ RHS.
