<s>[INST] Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$ Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$.
Find $(a+b+c)$.
I computed the whole product ;If  $(a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16$. Unable to view how to proceed further.
Please help.
 [/INST] Since scaling by a constant
leaves the product unchanged,
we can assume that
$a=1$.
This becomes
$(1+b+c)(1+1/b+1/c)
=16$.
Looking for a special solution,
assume $b = c$.
This becomes
$16
=(1+2b)(1+2/b)
=1+2b+2/b+4
$
or
$11
=2b+2/b
$
or
$2b^2-11b+2 = 0$.
Solving this,
$b 
=\dfrac{11\pm \sqrt{11^2-16}}{4}
=\dfrac{11\pm \sqrt{105}}{4}
$.
So these are two solutions.
Going back to 
the general case,
$16
=(1+b+c)(1+1/b+1/c)
=1+b+c+1/b+1/c+(b+c)(1/b+1/c)
=1+b+c+1/b+1/c+2+b/c+c/b
$
or
$13
=b+c+1/b+1/c+b/c+c/b
$.
If
$c = rb$,
this becomes
$13
=b+rb+1/b+1/(rb)+r+1/r
$
or
$13r
=br+r^2b+r/b+1/(b)+r^2+1
$
or
$(b+1)r^2+(b-13+1/b)r+1+1/b
=0
$.
The discriminant of this is
$d^2
=(b-13+1/b)^2-4(b+1)(1+1/b)
=(b^4 - 30 b^3 + 163 b^2 - 30 b + 1)/b^2
$
(according to Wolfy).
This has real roots
$b 
= \dfrac{7 \pm 3 \sqrt{5}}{2},
2/(23 + 5 \sqrt(21)),
23/2 + (5 \sqrt(21))/2
$
with approximate values
$0.14590,
6.8541,
0.043561,
22.956
$.
The plot shows that
for $b$ from
0.14590 to 6.8541
and $b > 22.956$
this is positive,
so there are real roots
for these $b$.
By taking the positive square root,
we can get a positive $r$.
I'll stop here.
</s>