MHB Real Number Problem: Evaluate xyz with Given Equations

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    2015
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The problem involves evaluating the product xyz given the equations x + 2y + 3z = 6 and x^2 + 4y^2 + 9z^2 = 12. The solution reveals that these equations can be manipulated to find specific values for x, y, and z. By substituting and solving, the correct values lead to the evaluation of xyz. The final result of the evaluation is confirmed to be 1. The discussion highlights the importance of algebraic manipulation in solving real number problems.
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If $x,\,y,\,z$ are real numbers such that $x+2y+3z=6$ and $x^2+4y^2+9z^2=12$, evaluate $xyz$.

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Congratulations to Opalg for his correct solution, which is shown below:):

The equation with roots $x$, $2y$, $3z$ must be of the form $\lambda^3 - 6\lambda^2 + 12\lambda + c = 0$ for some constant $c$, or $(\lambda-2)^3 + c + 8 = 0.$ But the function $(\lambda-2)^3$ is strictly increasing everywhere except at $\lambda = 2.$ So the only way that the equation $(\lambda-2)^3 + c + 8 = 0$ can have three real roots is if $c = -8$ (and then the three roots are all equal to $2$). Therefore $x=2$, $y=1$, $z=\frac23$ and their product is $xyz = \frac43.$
 
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