How can I prove BMO2 1995 Q3? It reads,
Let a, b, c be real numbers satisfying
1. a < b < c,
2. a+b+c = 6,
3. ab+bc+ca = 9.
Prove 0 ≤ a ≤ 1 ≤ b ≤ 3 ≤ c ≤ 4.
The key to answering this question is to sketch the graph of x3 - 6x2 + 9x + s = 0 for all values of s for which the polynomial has three roots.
Here's the motivation for this approach...
Let a, b, and c be the three real roots of f(x) = x3+qx2+rx+s.
From the facts given here (a+b+c=6, ab+bc+ca=9), we know q=-6, r=9.
So the polynomial is f(x) = x3 - 6x2 + 9x + s
The derivative of this is polynomial 3x2-12x+9 which has roots at 1 and 3, so sketching the graph, we see it has a relative maximum (whose value is s+4) at x=1 and a relative minimum (whose value is s) at x=3.
Did I mention "sketching the graph"? That's the key to understanding this problem!
From the sketch, it's easy to see that two of the roots must be on either side of the relative maximum and two of the roots must be on either side of the relative minimum. That fixes a ≤ 1 ≤ b ≤ 3 ≤ c.
Now, let's see about the endpoints...
Because the value of the relative maximum is s+4 and the value of the relative minimum is s, it follows that in order for this to have three roots, s must be at most 0 and at least -4. Remember -s=abc, and -s is nonnegative so either one of the roots is zero or else an even number of roots are negative. Since b and c are nonnegative, a must be nonnegative as well.
Finally, viewing c as a function of s on the interval -4 ≤ s ≤ 0, we see it's a decreasing function. A good way to understand this is: by decreasing s, we're lowering the entire graph, which causes the largest root to move to the right. That is, when s decreases, c increases. So the smallest possible value of s gives you the largest possible value of c. When s=-4, then c=4, which is the largest possible value of c.
So 0 ≤ a ≤ 1 ≤ b ≤ 3 ≤ c ≤ 4.
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