Exercise 2133Algebra › Equations and radicalsOlympiadAdded 3 Nov 2026

A nonlinear 3×3 system

#nonlinear system#olympiad#symmetry

Statement

Solve in $(\mathbb{R}_+)^3$ the system
$\begin{cases}x+y+z=3\\ x^2+y^2+z^2=3\\ xyz=1\end{cases}$

Hint

Show that $x=y=z=1$ using AM-GM.

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Solution

AM-GM. From $xyz=1$ and $x+y+z=3$ : by AM-GM, $x+y+z\ge3\sqrt[3]{xyz}=3$ , with equality iff $x=y=z$ . Since the first equation gives $x+y+z=3$ , equality must hold, so $x=y=z$ .

Then. $3x=3\Rightarrow x=1$ . Check the second equation: $3\cdot1=3$ . OK.

Conclusion. $(x,y,z)=(1,1,1)$ .

A nonlinear 3×3 system

Statement

Solution

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