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.
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 $(a+b+c)$
2
$\begingroup$
algebra-precalculus
inequality
systems-of-equations
a.m.-g.m.-inequality
cauchy-schwarz-inequality
-
1Don't you see it can't be done? There are just too many degrees of freedom. – 2017-02-27
-
0@YuriiSavchuk, this will give just $3a\cdot \frac3a = 9 \ne 16$ – 2017-02-27
-
5To be more specific, if $(a_0,b_0,c_0)$ is a solution, then so are $(2a_0, 2b_0, 2c_0)$ and $(1/a_0,1/b_0,1/c_0)$ and a lot of similar variations. Most of these triples do not have the same sum. – 2017-02-27
-
0Sorry that was wrong;edited @AndreiKulunchakov;@Arthur – 2017-02-27
-
0After the edit, if $(a_0,b_0,c_0)$ is a solution then $(1/a_0,1/b_0,1/c_0)$ is still a solution. – 2017-02-27
-
0After the edit, CS inequality will give the answer directly. – 2017-02-27
4 Answers
4
By $AM \ge GM$ inequality,$$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge \left(4\sqrt[4]{abc}\right)\left(4\sqrt[4]{\frac{1}{abc}}\right)=16$$and equality holds when $1=a=b=c$.
-
0Please explain equality part – 2018-07-08
-
0@CloudJR: that's part of the theorem - see e.g the [Wikipedia article](https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#The_inequality). – 2018-07-08
2
We know :$a>0 \to a+\frac1a \geq 2$ and now; $$(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\\ 1+1+1+1+a+b+c+\frac1a+\frac1b+\frac1c+\frac ab+\frac ac +\frac ba+\frac bc +\frac ca+\frac cb=\\ 4+(a+\frac 1a)+(b+\frac 1b)+(c+\frac 1c)+(\frac ba+\frac ab)+(\frac ac+\frac ca)+(\frac cb+\frac bc)\geq 4+3(2)+3(2)\\ l.h.s \geq 16$$ only $l.h.s=16$ taht $a=b=c=1$so $$a+b+c=1+1+1=3$$
-1
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.
-
0I have edited it;sorry for the inconvenience – 2017-02-27
-
0Note for the unwary (like me): this answer seems crazy, but that is because it tried to answer the OP's original question which was wrong and has now been edited. I hate it when that happens. – 2018-07-08
-2
By applying Caushy Schwartz inequality we get ( 1 + a + b + c ) ( 1 + 1/a + 1/b + 1/c ) >= 16 Equality holds when a^2 = b^2 = c^2 = 1 Therefore a = b = c = 1