greatest integer

[man111]

Show that [tex]\left[5x\right]+\left[5y\right]\geq \left[3x+y\right]+\left[3y+x\right][/tex]

Where [tex]x,y\geq 0[/tex] and [tex]\left[.\right]=[/tex] greatest integer function

[martin123456]

If [tex]x\ge 1[/tex] denote [tex]z= x-1 \ge 0[/tex]. Then the RHS is [tex][5z+5]+[5y]=[5z]+[5y]+5[/tex] and the LHS [tex][3z+y+3]+[3y+z+1]=[3z+y]+[3y+z]+4[/tex]. So the inequality becomes [tex][5z]+[5y]+1 \ge [3z+y]+[z+3y][/tex] which is weaker than the stated.
So WLOG we assume that [tex]0 \le x\le y <1[/tex].
[tex]3x+y \le 5x \Leftrightarrow y \le 2x[/tex], [tex]x+3y \le 5y \Leftrightarrow x \le 2y[/tex]. But [tex]x \le y \Rightarrow x \le 2y[/tex]. So when this holds we have [tex][3x+y]\le [5x][/tex] and [tex][y+3x] \le [5y][/tex]. So the inequality holds.
Now assume [tex]y > 2x[/tex].
(1) [tex]y \in [0, \frac{1}{5}][/tex].Then [tex]x < \frac{1}{10}[/tex]. So [tex]3x+y < \frac{3}{10}+\frac{1}{5}=\frac{1}{2}<1[/tex] so [tex][tex][3x+y]=0[/tex]. Since [tex]3x+y \le x+3y[/tex] we have [tex][x+3y]=0[/tex] too. So the inequality holds.
(2) [tex]y \in (\frac{1}{5}, \frac{2}{5}][/tex]. So [tex]3x+y\le \frac{3}{5}+\frac{2}{5}=1[/tex]. If we dont have equality then the case is like in (1). If an equality holds than we can substitude [tex]y=\frac{2}{5}[/tex] and [tex]x=\frac{1}{5}[/tex] to get [tex]1+2 \ge 1+1[/tex].
(3) [tex]y \in (\frac{2}{5},\frac{3}{5}][/tex]. Then [tex]x < \frac{3}{10}[/tex].
(3.1) if [tex]x < \frac{1}{5}[/tex] then [tex][5x]=0[/tex], [tex][5y][/tex] is 2 or 3, [tex]\frac{2}{5}<3x+y\le \frac{6}{5} \Rightarrow [3x+y][/tex] is 0 or 1 and [tex][x+3y][/tex] is 1. So LHS is max 2, LHS is greater or equal to 2.
(3.2) [tex]x=\frac{1}{5}[/tex] then we have [tex]1+[5y] \ge [\frac{3}{5}+y]+[\frac{1}{5}+y][/tex]. LHS is 1+0=1, RHS is at least 1.
(3.3) [tex]x > \frac{1}{5}[/tex] so [tex][5x]=1[/tex], [tex][5y][/tex] is 2 or 3, so RHS is 3 or 4. [tex]1<3x+y <\frac{9}{10}+\frac{6}{10}=\frac{3}{2}[/tex], so the floor has value of 1. We know that [tex]3x+y<x+3y[/tex] so LHS is less than 2.
(4) [tex]y \in (\frac{3}{5}, \frac{4}{5}][/tex]. Then [tex]x\le \frac{2}{5}[/tex], so [tex]\frac{4}{5}\le3x+y\le 2[/tex] so it's 0 or 1 or 2 and [tex]x+3y\le\frac{6}{5}[/tex] so it's 0 or 1. So LHS is at most 3. RHS: [tex][5y][/tex]=3 or 4.
(5)[tex]y \in (\frac{4}{5},1)[/tex]. Then [tex]x<\frac{1}{2}[/tex]. [tex]\frac{4}{5} <3x+y\le \frac{3}{2}+1=\frac{5}{2}=2.5[/tex]. So it's 0 or 1 or 2. [tex]x+3y[/tex] is less than [tex]\frac{1}{2}+\frac{12}{5}=\frac{5+24}{10}=2.9[/tex] and more than 2 so the floor is 2. So the max value of the LHS is 4. RHS: [5y] is 4 or 5.