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

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The discussion focuses on maximizing the expression $a+b+c$ under the constraints $a \in \mathbb{Z}$, $b, c \in \mathbb{R}$, with the conditions $a > b$ and $a > c$. The equations provided are $a + 2b + 3c = 6$ and $abc = 5$. The goal is to determine the maximum value of $a + b + c$ given that the minimum value of $a$ is defined as $k$. The analysis leads to specific values for $a$, $b$, and $c$ that satisfy the equations while adhering to the constraints.

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

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 :$(a,b,c)=(10,\dfrac {-3}{2}, \dfrac {-1}{3})$
$\therefore max(a+b+c)=10-\dfrac {1}{2}-1=\dfrac{17}{2}$
 
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