Finding $max(a+b+c)$ Given $a\in Z, b,c\in R$

[Albert1]

$a\in Z,\,\, b,c\in R$
$a>b ,\,\, and ,\, a>c$
given:
$a+2b+3c=6---(1)$
$abc=5---(2)$
if :$min(a)=k$
find: $max(a+b+c)$

 

[Albert1]

$a\in Z,\,\, b,c\in R$
$a>b ,\,\, and ,\, a>c$
given:
$a+2b+3c=6---(1)$
$abc=5---(2)$
if :$min(a)=k$
find: $max(a+b+c)$

Spoiler

from (1) and (2) we know :$a\in N$
$c=\dfrac {6-2b-k}{3}=\dfrac {5}{kb}$
we have:$2kb^2+b(k^2-6k)+15=0----(*)$
if $b \in R $ then :
$(k^2-6k)^2\geq120k$
or:$k(k-6)^2\geq 120$
$\therefore min(a)=k=10$
from (*):$4b^2+8b+3=0 $
and $b=\dfrac {-1}{2} or,\, \dfrac {-3}{2}$
and :$(a,b,c)=(10,\dfrac {-1}{2}, -1)$
or :$(a,b,c)=(10,\dfrac {-3}{2}, \dfrac {-1}{3})$
$\therefore max(a+b+c)=10-\dfrac {1}{2}-1=\dfrac{17}{2}$