1988 USAMO Problems/Problem 2
Problem
The cubic polynomial has real coefficients and three real roots . Show that and that .
Solution
Solution 1
By Vieta's Formulas, , , and . Now we know ; in terms of r, s, and t, then, Now notice that we can multiply both sides by 2, and rearrange terms to get . But since , the three terms of the RHS are all non-negative (as the square of a real number is always non-negative), and therefore their sum is also non-negative -- that is, .
Now, we will show that . We can square both sides, and the inequality will hold since they are both non-negative (it is given that , therefore ). This gives . Now we already have , so substituting this for k gives Note that this is a quadratic. Since its leading coefficient is positive, its value is less than 0 when s is between the two roots. Using the quadratic formula: The quadratic is 0 when s is equal to r or t, and the inequality holds when its value is less than or equal to 0 -- that is, . (Its value is less than or equal to 0 when s is between the roots, since the graph of the quadratic opens upward.) In fact, the problem tells us this is true. Q.E.D.
Solution 2
From Vieta's Formula (which tells us that and ), we have that clearly non-negative. To prove , it suffices to prove the square of this relation, or This in turn simplifies to or which is clearly true as . This completes the proof.
Solution 3
By Vieta's Formulas, and . .
To show that , simply note that by the trivial inequality, all three squares are greater than as they are the squares of real numbers.
To show that , since both are positive, it is sufficient to show that . implies that . . Let and . We then have , which is clearly true as both and are positive.
See Also
1988 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.