<s>[INST] Prove inequality $a<1Given $a,b,c$ 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
 [/INST] 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 $(c, \infty)$.
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