1. ## Tough Problem from an olympiad

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

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

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

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

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

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

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

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

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

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

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

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

5. Thanks NCA

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

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

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

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