Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Prove it United States

Problem:

Suppose xx, yy, and zz are real numbers that satisfy x+y+z>0x + y + z > 0, xy+yz+zx>0x y + y z + z x > 0 and xyz>0x y z > 0. Prove that xx, yy, and zz must all be positive.

Solution

Solution:

Note that xx, yy, and zz are the roots of the polynomial
(tx)(ty)(tz)=t3(x+y+z)t2+(xy+xz+yz)txyz. (t - x)(t - y)(t - z) = t^{3} - (x + y + z) t^{2} + (x y + x z + y z) t - x y z.
We claim that this polynomial has no negative roots. To see this, if we plug in a negative value of tt, all four terms we are adding are negative, so the sum is negative. If we plug in t=0t = 0, the first three terms are negative, and xyz-x y z is negative since xyzx y z is positive, so the sum is again negative. Thus, whenever t0t \leq 0, the sum is negative, so any real root must satisfy t>0t > 0. So, since xx, yy, and zz are all roots, they must all be positive.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.