Question

If $a,b,c$ are integers with $c>0$ the prove $\left[\frac{2a}{c}\right]+\left[\frac{2b+1}{c}\right]\ge \left[\frac{a}{c}\right]+\left[\frac{b}{c}\right]+\left[\frac{a+b+1}{c}\right]$

Answer 1

Let $a=mc+r$ and $b=nc+s$

where m, n, r, s are integer and $0\le r<cand0\le s<c$

Now, we will establish first on concept if $s\ge r,$ then $2s+1\ge r+s+1$

$[\frac{2s+1}{c}\ge \left[\frac{r+s+1}{c}\right]]$ .....(1)

If s< r. then $2r>r+s+1$

and $\left[\frac{2r}{c}\right]\ge \left[\frac{r+s+1}{c}\right]$.........(2)

From (1) and (2)

$\left[\frac{2r}{c}\right]+\left[\frac{2s+1}{c}\right]\ge \left[\frac{r+s+1}{c}\right]$

Since, $\left[\frac{r}{c}\right]=\left[\frac{s}{c}\right]=0$

we can also say that

$\left[\frac{2r}{c}\right]+\left[\frac{2s+1}{c}\right]\ge \left[\frac{r}{c}\right]+\left[\frac{s}{c}\right]+\left[\frac{r+s+1}{c}\right]$ .........(3)

Now $\left[\frac{2a}{c}\right]+\left[\frac{2b+1}{c}\right]=[2m+\frac{2r}{c}]+[2n+\frac{2s+1}{c}]$

$=2m+2n+\left[\frac{2r}{c}\right]+\left[\frac{2s+1}{c}\right]\ge 2m+2n+\left[\frac{r}{c}\right]+\left[\frac{r+s+1}{c}\right]$

(from (3))$\ge m+\left[\frac{r}{c}\right]+n+\left[\frac{s}{c}\right]+m+n+\left[\frac{r+s+1}{c}\right]$

$\ge [m+\frac{r}{c}]+[n+\frac{s}{c}]+[m+n+\frac{r+s+1}{c}]$

$\ge \left[\frac{a}{c}\right]+\left[\frac{b}{c}\right]+\left[\frac{a+b+1}{c}\right]$

$\therefore \left[\frac{2a}{c}\right]+\left[\frac{2b+1}{c}\right]\ge \left[\frac{a}{c}\right]+\left[\frac{b}{c}\right]+\left[\frac{a+b+1}{c}\right]$

Hence proved.