Let $ a,b,c$ be three pairwise distinct real numbers such that $ a+b+c=6=ab+bc+ca-3$. Prove that $ 0<abc<4$.
Problem
Source: oliforum contest, round 1, problem 3
Tags: calculus, derivative, symmetry, inequalities unsolved, inequalities
Kunihiko_Chikaya
22.09.2008 04:11
The condition for which $ y = x(x - 3)^2$ has 3 distinct intersection of points with $ y = abc$ gives $ 0 < abc < 4$. In detail http://www.mathlinks.ro/Forum/viewtopic.php?t=207499
me@home
22.09.2008 04:48
Suppose we have three real $ a,b,c$ that solve the equation. That means that $ x^3-6x^2+9x-abc=0$ has the three roots $ a,b,c$ which are all real.
Thus the line $ y=abc$ meets $ x(x-3)^2$ in three points. Thus $ abc$ lies (vertically) between the two zeros of the derivative of the cubic.
\[ \frac{d}{dx}x(x-3)^2 = (x-3)^2+2x(x-3) = 3(x-3)(x-1) = 0\]
Thus $ abc\in( f(1),f(3) )$, or $ (0,4)$
Kunihiko_Chikaya
22.09.2008 04:55
me@home, have you seen the link that I gave? Some Japanese high school student could solve the problem in 30 seconds.
Kunihiko_Chikaya
22.09.2008 05:14
More difficult problem. What's the range of $ a^3 + b^3 + c^3?$
NickNafplio
22.09.2008 07:27
kunny wrote: More difficult problem. What's the range of $ a^3 + b^3 + c^3?$
$ a^3 + b^3 + c^3 = (a + b + c)^3 - 3(a + b)(b + c)(c + a) = (a + b + c)^3 - 3((a + b + c)(ab + bc + ca) - abc) = ... = 54 + 3abc$
Hence $ 54 < a^3 + b^3 + c^3 < 66$
Kunihiko_Chikaya
22.09.2008 07:29
That's right.
Kamyar1991
22.09.2008 13:15
i have simple solution by delta. $ c=6-(a+b)$ , $ c(a+b)+ab=9$ , $ a^2+a(b-6)+9-6b=0$ now we get delta: $ \Delta = (b-6)^2-4(b-3)^2>0$ so we get $ \abs (b-6)>abs(2b-6)$ if $ b>3$ then $ 6-b>6-2b$ , so $ b>0$ then by symmetry $ a,b,c>0$ then $ abc>0$ between $ ab,ac,bc$ we know on of them is must greater than 1 like $ ab>1$ we consider $ ab=y>1$ and $ x=a+b$ by $ AM-GM$ ineq $ x>=2\sqrt{y}$ and we knew $ (6-x)(x)+y=9$ $ x^2-6x+9-y=0$ so the solutions are $ x_1,x_2=3\pm \sqrt{y}$ if $ x=3-\sqrt{y}$ then at before we knew $ x>=2\sqrt{y}$ then $ y<1$ contradiction. thus $ x=3+\sqrt{y}$ now we must prove that $ abc<4$ it means that $ y(6-x)<4$ $ y(3-\sqrt{y})<4$ we know $ y>0$ exist $ s=\sqrt{y}>1$ now our purpose is show that $ 3s^2<4+s^3$ we know $ s>1$ then we consider $ s=t+1$ $ 3(t+1)^2<4+(t+1)^3$ it means that $ t^3+2>3t$ this is obviouse by $ AM-GM$ ineq. then we are done.
NickNafplio
22.09.2008 19:03
Here is the solution I sent for the contest
We have:
$ ab + bc + ca = 9 \Leftrightarrow ab = 9 - c(a + b) = 9 - c(6 - c) = c^2 + 9 - 6c = (c - 3)^2 \Leftrightarrow abc = c(c - 3)^2$
Similarly we get: $ abc = b(b - 3)^2$, and $ abc = a(a - 3)^2$
Hence $ a(a - 3)^2 = b(b - 3)^2 = c(c - 3)^2 = abc$
Suppose that $ abc = 0$, then two of the three numbers must be equal with $ 0$ or $ 3$, but this is a contradiction because the three numbers are distinct, hence we have:
$ abc = a(a - 3)^2 = b(b - 3)^2 = c(c - 3)^2 \neq 0$.
Now obviously the three numbers are not equal with $ 0$ or $ 3$ and all three have the same sign, and because $ a + b + c = 6 > 0$
we have that all three numbers are positive, hence $ abc > 0$
Suppose now that:
$ abc > 4 \Leftrightarrow abc - 4 = a(a - 3)^2 - 4 = b(b - 3)^2 - 4 = c(c - 3)^2 - 4 > 0 \Leftrightarrow c^3 - 6c^2 + 9c - 4 = a^3 - 6a^2 + 9a - 4 = b^3 - 6b^2 + 9b - 4 > 0 \Leftrightarrow (c - 1)^2(c - 4) = (b - 1)^2(b - 4) = (c - 1)^2(c - 4) > 0 \Leftrightarrow a,b,c > 4 \Rightarrow a + b + c = 6 > 12$,
which is a contradiction. Hence we have $ abc \leq 4$
I shall prove now that $ abc \neq 4$. Suppose that:
$ abc = 4 \Leftrightarrow abc - 4 = a(a - 3)^2 - 4 = b(b - 3)^2 - 4 = c(c - 3)^2 - 4 = 0 \Leftrightarrow c^3 - 6c^2 + 9c - 4 = a^3 - 6a^2 + 9a - 4 = b^3 - 6b^2 + 9b - 4 = 0 \Leftrightarrow (c - 1)^2(c - 4) = (b - 1)^2(b - 4) = (c - 1)^2(c - 4) = 0$,
which is true only if two of the three numbers are equal with $ 1$ or $ 4$, but this is again a contradiction because the three numbers are distinct. Hence $ abc \neq 4$
Hence we have $ 0 < abc < 4$