# Thread: Tough Problem from an olympiad

1. ## Tough Problem from an olympiad

Let $\displaystyle a<b<c$ be real numbers such that $\displaystyle a+b+c=6$ and $\displaystyle ab+bc+ca=9$ then prove that $\displaystyle 0<a<1<b<3<c<4$.
Also,find the range of $\displaystyle a^3+b^3+c^3$.

2. Hello Sir,
I haven't solved the problem

but I think this can help a bit for the second part
a+b+c=6
b+bc+ca=9
(a+b+c)^2 = 36
a^2+b^2+c^2 +2(ab+bc+ca)=36
a^2+b^2+c^2=18
now
a^3+b^3+c^3 =(a+b+c)(a^2+b^2+c^2-ab-bc-ca) +3abc
=6(18-9)+3abc
=54+3abc

I stopped here

But I think
using a cubic equation like
x^3+6x^2+9x+m=0
can help
where m=abc

3. Originally Posted by pankaj

Let $\displaystyle a<b<c$ be real numbers such that $\displaystyle a+b+c=6$ and $\displaystyle ab+bc+ca=9$ then prove that $\displaystyle 0<a<1<b<3<c<4.$
so $\displaystyle a,b,c$ are roots of: $\displaystyle x^3-6x^2 + 9x -abc=0,$ which gives us: $\displaystyle (a-3)^2=bc, \ (b-3)^2=ac, \ (c-3)^2=ab.$ call this (1). note that all $\displaystyle a,b,c$ are non-zero, because otherwise we would have

$\displaystyle abc=0$ and thus $\displaystyle 0=x^3-6x^2+9x=x(x-3)^2,$ which contradicts that $\displaystyle a,b,c$ are pairwise distinct. now by (1) either all $\displaystyle a,b,c$ are positive or all of them are negative. but since $\displaystyle a+b+c >0,$

we must have that all $\displaystyle a,b,c$ are positive. also since $\displaystyle a+b+c=6$ and $\displaystyle a<b<c,$ we have $\displaystyle a < 3.$ hence by (1): $\displaystyle a=3-\sqrt{bc}$ and thus: $\displaystyle 6=a+b+c=b-\sqrt{bc}+c+3.$ that will give us

$\displaystyle b-\sqrt{bc} + c - 3=0.$ call this (2). the discriminant of $\displaystyle b -\sqrt{bc} + c - 3,$ as a quadratic function in $\displaystyle \sqrt{b},$ is $\displaystyle 12 - 3c,$ which must be positive. therefore $\displaystyle \boxed{c < 4}.$ also since $\displaystyle b < c$ and by (2) we have

$\displaystyle \sqrt{b}(\sqrt{b}-\sqrt{c})=3-c,$ and $\displaystyle \sqrt{c}(\sqrt{c}-\sqrt{b})=3-b,$ we must have $\displaystyle \boxed{c > 3}$ and $\displaystyle \boxed{b < 3}.$ finally since c > 3, we'll get from (1) that $\displaystyle c=3+\sqrt{ab}.$ thus: $\displaystyle 6=a+b+c=a+b+\sqrt{ab}+3.$ hence

$\displaystyle a+b+\sqrt{ab}=3.$ but: $\displaystyle 3a < a+b+\sqrt{ab} < 3b,$ because $\displaystyle a<b.$ thus $\displaystyle \boxed{a< 1}$ and $\displaystyle \boxed{b>1}$ and we are done.

Also,find the range of $\displaystyle a^3+b^3+c^3.$
since $\displaystyle abc=a(a-3)^2$ and $\displaystyle 0 < a < 1,$ it's clear that $\displaystyle 0 < abc < 4.$ but this is not enough to claim that the range of $\displaystyle abc$ is $\displaystyle (0,4)$ because it might be smaller. in fact we'll prove that the range is

indeed the interval $\displaystyle (0,4)$: let $\displaystyle abc=k.$ the cubic $\displaystyle x^3 -6x^2+9x - k=0$ must have 3 distinct real roots $\displaystyle a,b,c,$ which happens iff $\displaystyle \Delta,$ the discriminant of the cubic, is positive. a simple calculation

shows that $\displaystyle \Delta=27k(4-k). \ \Box$

4. ok, the second part of the question was to find the range of $\displaystyle a^3 + b^3 + c^3$ and, for some strange reasons, i thought it was to find the range of $\displaystyle abc.$ however, since $\displaystyle a^3+b^3+c^3=54+3abc,$

as ADARSH showed, and the range of $\displaystyle abc,$ as i showed, is (0,4), the range of $\displaystyle a^3+b^3+c^3$ will be $\displaystyle (54, 66).$

5. Thanks NCA

The value of $\displaystyle abc$ can also be found by the fact that if $\displaystyle f(x)=0$ will have $\displaystyle 3$ roots then $\displaystyle f(1)f(3)<0$ where $\displaystyle x=1$ and $\displaystyle x=3$ are roots of $\displaystyle f'(x)=0$

6. Hello Adarsh and NCA there is also a Caculus approach to this problem.

You see there are two critical points of $\displaystyle f(x)=0$,$\displaystyle viz.x=1,3.$
Since $\displaystyle f(x)$ is a cubic($\displaystyle f(x)=(x-a)(x-b)(x-c)=x^3-6x^2+9x-abc)$ with coefficient of $\displaystyle x^3$ positve it is obvious that $\displaystyle x=1$ is a point of maxima and $\displaystyle x=3$ is a point of minima(can also be ascertained by the double derivative test).
$\displaystyle f(0)=-abc$
$\displaystyle f(1)=4-abc$
$\displaystyle f(3)=-abc$
$\displaystyle f(4)=4-abc$
As I said before that for $\displaystyle f(x)=0$ to have $\displaystyle 3$ roots $\displaystyle f(1)f(3)<0$ which gives $\displaystyle 0<abc<4$.
Now it is obvious that there is one root between $\displaystyle x=0$ and $\displaystyle x=1$,another root between $\displaystyle x=1$ and $\displaystyle x=3$ and the third root between $\displaystyle x=3$ and $\displaystyle x=4$

7. Originally Posted by pankaj
Hello Adarsh and NCA there is also a Caculus approach to this problem.

You see there are two critical points of $\displaystyle f(x)=0$,$\displaystyle viz.x=1,3.$
Since $\displaystyle f(x)$ is a cubic with coefficient of $\displaystyle x^3$ positve it is obvious that $\displaystyle x=1$ is a point of maxima and $\displaystyle x=3$ is a point of minima(can also be ascertained by the double derivative test).
$\displaystyle f(0)=-abc$
$\displaystyle f(1)=4-abc$
$\displaystyle f(3)=-abc$
$\displaystyle f(4)=4-abc$
As I said before that for $\displaystyle f(x)=0$ to have $\displaystyle 3$ roots $\displaystyle f(1)f(3)<0$ which gives $\displaystyle 0<abc<4$.
Now it is obvious that there is one root between $\displaystyle x=0$ and $\displaystyle x=1$,another root between $\displaystyle x=1$ and $\displaystyle x=3$ and the third root between $\displaystyle x=3$ and $\displaystyle x=4$
this solution, although not very accurate in this form, will work. the reason that i didn't use calculus was because you posted the question in the pre-algebra subforum! so i thought you were

looking for an elementary solution.

8. Question was posted a long time ago.How ADARSH retrieved it,well that goes to his credit.
In fact I was looking for an algebraical solution which you have provided.
The method that I have pointed out is not genral one and may work for this specific problem.