Given $a,b,c$ are positive number satisfy $a<b<c;a+b+c=6;ab+bc+ca=9$. Prove that $a<1<b<3<c<4$
i think use quality:"The solutions of function $f(x)=(x-a)(x-b)(x-c)$ satisfy $a\le x_1\le b\le x_2\le c$ with $x_1,x_2\ (x_1<x_2)$ are extreme point of $f(x)$ ".
Help me
Your idea is correct. $a, b, c$ are the zeros of $$ f(x)=(x-a)(x-b)(x-c) = x^3 - 6 x^2 + 9x - abc $$ Rolles's theorem states that $f'$ has a zero in each interval $(a, b)$ and $(b, c)$. But $$ f'(x) = 3x^2 - 12 x + 9 = 3(x-1)(x-3) $$ has zeros $x_1=1$ and $x_2=3$. It follows that $$ a < x_1 = 1 < b < x_2 = 3 < c \, . $$
It remains to show that $c < 4$: $f$ changes signs exactly at the zeros $a, b, c$, and $f(4) = 4 - abc = f(1)$. Therefore $4$ must lie in the interval $(c, \infty)$.
Let $abc=w^3$.
By the Martin's idea we see that $a$, $b$ and $c$ are roots of the equation
$$w^3=f(x),$$ where $f(x)=x(x-3)^2$.
Draw a graph of $f$ and graph of the line $y=w^3$,
which intersects the graph of $f$ in three following points
$A(a,f(a))$, $B(b,f(b))$ and $C(c,f(c))$.
Since $$f'(x)=3(x-1)(x-3),$$ we see that $0<a<x_{max}<b<x_{min}<c$.
But $x_{max}=1$ and $x_{min}=3$, which gives $0<a<1<b<3<c$.
In another hand $c$ gets a maximal value, when $w^3$ gets a maximal value,
which happens for $a\rightarrow b\rightarrow1$, which gives $c<4$ and we are done!
If you expand the function $f(x)$, you would get a parametric polynomial. Some of its coefficients are given in the question. Just plug them in and then your polynomial would only have one unknown coefficient. Then, take the derivative and find local extrema of your function. You would find out the extrema do not depend on the unknown coefficient and they are fixed numbers. As $a$ is the smallest root of $f(x)$, it cannot be greater than the first extremum. Because, if $a$ is larger than or equal to the first extrema, then we would not have all the roots (we can have a maximum of two roots). The same reasoning can be applied to the rest. consider a wrong case and proof it is not true.
