Let be real numbers such that and then prove that .
Also,find the range of .
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
so are roots of: which gives us: call this (1). note that all are non-zero, because otherwise we would have
and thus which contradicts that are pairwise distinct. now by (1) either all are positive or all of them are negative. but since
we must have that all are positive. also since and we have hence by (1): and thus: that will give us
call this (2). the discriminant of as a quadratic function in is which must be positive. therefore also since and by (2) we have
and we must have and finally since c > 3, we'll get from (1) that thus: hence
but: because thus and and we are done.
since and it's clear that but this is not enough to claim that the range of is because it might be smaller. in fact we'll prove that the range isAlso,find the range of
indeed the interval : let the cubic must have 3 distinct real roots which happens iff the discriminant of the cubic, is positive. a simple calculation
shows that
Hello Adarsh and NCA there is also a Caculus approach to this problem.
You see there are two critical points of ,
Since is a cubic( with coefficient of positve it is obvious that is a point of maxima and is a point of minima(can also be ascertained by the double derivative test).
As I said before that for to have roots which gives .
Now it is obvious that there is one root between and ,another root between and and the third root between and
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.