Matematické Fórum

stuart clark · 10. 01. 2012 09:22

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

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

vanok · 22. 01. 2012 15:35 — Editoval vanok (22. 01. 2012 16:01)

Hi ↑ stuart clark:,

To resolve this problem, I use that $x = E (x) + \{x \}$ , where $\{x \}$ is the fractional part of $x$ ; $E(.)$ is greatest integer function
$y = E ( y ) + \{y\}$
and
$E( x+n ) =E(x) +n$ , where $n$ is an integer.

The relation to be verified:
$E(5x)+ E(5y) \ge E(3x+y)+ E( x+3y)$
becomes
$E(x)+E(y)+E(5 \{x \})+E(5 \{y\})\ge E (3 \{ x \}+ \{y \}) + E ( \{ x \}+3 \{y \})$
As, it is possible that $x \in [0; 1 [$ , we have to demonstrate
$E(5 \{x \})+E(5 \{y\})\ge E (3 \{ x \}+ \{y \}) + E ( \{ x \}+3 \{y \})$

To establish this relation, I show that a number finished by check is enough.

I made everything on board a graphic representation $E(5x)+ E(5y)= k$ ,
where k is a natural number.
For example $E(5x)+ E(5y)= 4$ , is formed by rosary of the intrerior of squared, including his edge, with the exception of the vertice the most distant from origin, aside $\frac15$ among which two vertices are situated on the right of equation $x+y= \frac45$ , the coordinates of these vertices, on this right are $(0; \frac 45); (\frac 15; \frac 35); (\frac 25; \frac 25); (\frac 35; \frac 15); (\frac 45; 1)$

An expression as $3x+y$ or $x+3y$ take the maximal value in the vertice the most remote from origin of a such squared.
For example in the squared of coordinates of which are $(\frac 25; \frac 15); (\frac 35; \frac 15); (\frac 35; \frac 25); (\frac 25; \frac 25)$ it is about vertice of coordinates $( \frac 35; \frac 25)$
Held account that the function $E(5 \{x \})+E(5 \{y\})\ge E (3 \{ x \}+ \{y \}) + E ( \{ x \}+3 \{y \})$ E also possesses points discontinuity, I study
$E (3 \{ x \}+ \{y \}) + E ( \{ x \}+3 \{y \})$
in the neighborhood of this vertice.

Number of checks can be reduced, because the formula to be demonstrated is symmetric in x and y.

After 15 check, I can assert that
$E(5x)+ E(5y) \ge E(3x+y)+ E( x+3y)$
is the valid formula.

stuart clark

↑ vanok:

Thanks vanok

actually i have face a problem in solving Interior points. now got it

Thanks vanok

Pavel · 22. 01. 2012 22:45

I do not understand the way of getting $E(x)+E(y)+E(5 \{x \})+E(5 \{y\})\ge E (3 \{ x \}+ \{y \}) + E ( \{ x \}+3 \{y \})$ from $E(5x)+ E(5y) \ge E(3x+y)+ E( x+3y)$ .

vanok · 22. 01. 2012 23:25 — Editoval vanok (25. 01. 2012 12:06)

Hi ↑ Pavel:,
We replace in
$E(5x)+ E(5y) \ge E(3x+y)+ E( x+3y)$ ... $(1)$

$E(5x)+ E(5y)$
By an expression which is equal to it
$E(5( E (x) + \{x \} ))+E(5( E (y) + \{y\}) )=$
$5( E (x)) +5( E (y))+E(5 \{x \})+E(5 \{y\})$
And we replace similarly
$E(3x+y);E( x+3y)$ in $(1)$
after the supression of the identical terms $4E(x)+4E(y)$ to the left and to the right

We obtain
$E(x)+E(y)+E(5 \{x \})+E(5 \{y\})\ge E (3 \{ x \}+ \{y \}) + E ( \{ x \}+3 \{y \})$