# Minimum value of a floor function [closed]

Minimum value of a floor function. $$\left \lfloor{(a+b+c)/d}\right \rfloor + \left \lfloor{(a+b+d)/c}\right \rfloor +\left \lfloor{(a+d+c)/b}\right \rfloor + \left \lfloor{(d+b+c)/a}\right \rfloor$$ My working: For a=b= -d= -c, I'm getting -4 as the answer. Is there a concrete method to find the minimum value?

## closed as off-topic by Jean-Claude Arbaut, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Math1000, max_zornMay 2 at 1:27

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• Please encode $\LaTeX$ within dollar signs $\$\$ and include your workings. – Mohammad Zuhair Khan May 1 at 20:04

If a,b,c,d are positive: By $$\lfloor x\rfloor>x-1$$, the function $$f= \lfloor{(a+b+c)/d} \rfloor + \lfloor{(a+b+d)/c} \rfloor + \lfloor{(a+d+c)/b} \rfloor + \lfloor{(d+b+c)/a} \rfloor>(a+b+c)/d+(a+b+d)/c+(a+d+c)/b+(d+b+c)/a-4=(a+b+c+d)/d+(a+b+c+d)/c+(a+b+c+d)/b+(a+b+c+d)/a-8=(a+b+c+d)(1/a+1/b+1/c+1/d)-8$$ (By Cauchy-Schwarz) $$\ge (a\cdot\frac{1}{a}+b\cdot\frac{1}{b}+c\cdot\frac{1}{c}+d\cdot\frac{1}{d})^2-8=8$$.
As an integer, $$f>8$$ means $$f\ge 9$$. When a=b=c=1.1 and d=1, the function f=9, hence the Minimum value of the floor function is 9.
If we allow a,b,c,d to be negative: Let $$a=b=c=-1,d=\frac{1}{n}$$ where $$n\ge 2$$ is any positive integer, then $$f=3-3n$$ In this case, $$f$$ can be sufficiently small.