1988 USAMO Problems/Problem 2
The cubic polynomial has real coefficients and three real roots . Show that and that .
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.
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.
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.
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