Maths Olympiad Prep

Library / /2 of 24

, 2003

Algebra Difficulty 5.3 AIME, harder Prove it Canada

Problem:
Find all real positive solutions (if any) to
x3+y3+z3=x+y+z, and x2+y2+z2=xyz. \begin{gathered} x^{3}+y^{3}+z^{3}=x+y+z, \text{ and } \\ x^{2}+y^{2}+z^{2}=x y z . \end{gathered}

Solutions — 3

Solution 1

Solution:
Let f(x,y,z)=(x3x)+(y3y)+(z3z)f(x, y, z)=\left(x^{3}-x\right)+\left(y^{3}-y\right)+\left(z^{3}-z\right). The first equation above is equivalent to f(x,y,z)=0f(x, y, z)=0. If x,y,z1x, y, z \geq 1, then f(x,y,z)0f(x, y, z) \geq 0 with equality only if x=y=z=1x=y=z=1. But if x=y=z=1x=y=z=1, then the second equation is not satisfied. So in any solution to the system of equations, at least one of the variables is less than 1. Without loss of generality, suppose that x<1x<1. Then
x2+y2+z2>y2+z22yz>yz>xyz. x^{2}+y^{2}+z^{2}>y^{2}+z^{2} \geq 2 y z>y z>x y z .
Therefore the system has no real positive solutions.

Solution 2

Solution:
We will show that the system has no real positive solution. Assume otherwise.
The second equation can be written x2(yz)x+(y2+z2)x^{2}-(y z) x+\left(y^{2}+z^{2}\right). Since this quadratic in xx has a real solution by hypothesis, its discriminant is nonnegative. Hence
y2z24y24z20 y^{2} z^{2}-4 y^{2}-4 z^{2} \geq 0
Dividing through by 4y2z24 y^{2} z^{2} yields
141y2+1z21y2 \frac{1}{4} \geq \frac{1}{y^{2}}+\frac{1}{z^{2}} \geq \frac{1}{y^{2}}
Hence y24y^{2} \geq 4 and so y2y \geq 2, yy being positive. A similar argument yields x,y,z2x, y, z \geq 2. But the first equation can be written as
x(x21)+y(y21)+z(z21)=0 x\left(x^{2}-1\right)+y\left(y^{2}-1\right)+z\left(z^{2}-1\right)=0
contradicting x,y,z2x, y, z \geq 2. Hence, a real positive solution cannot exist.

Solution 3

Solution:
Applying the arithmetic-geometric mean inequality and the Power Mean Inequalities to x,y,zx, y, z we have
xyz3x+y+z3x2+y2+z23x3+y3+z333 \sqrt[3]{x y z} \leq \frac{x+y+z}{3} \leq \sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}} \leq \sqrt[3]{\frac{x^{3}+y^{3}+z^{3}}{3}}
Letting S=x+y+z=x3+y3+z3S=x+y+z=x^{3}+y^{3}+z^{3} and P=xyz=x2+y2+z2P=x y z=x^{2}+y^{2}+z^{2}, this inequality can be written
P3S3P3S33 \sqrt[3]{P} \leq \frac{S}{3} \leq \sqrt{\frac{P}{3}} \leq \sqrt[3]{\frac{S}{3}}
Now P3P3\sqrt[3]{P} \leq \sqrt{\frac{P}{3}} implies P2P3/27P^{2} \leq P^{3} / 27, so P27P \geq 27. Also S3S33\frac{S}{3} \leq \sqrt[3]{\frac{S}{3}} implies S3/27S/3S^{3} / 27 \leq S / 3,
so S3S \leq 3. But then P33\sqrt[3]{P} \geq 3 and S331\sqrt[3]{\frac{S}{3}} \leq 1 which is inconsistent with P3S33\sqrt[3]{P} \leq \sqrt[3]{\frac{S}{3}}. Therefore the system cannot have a real positive solution.

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