MHB Do $x, y, z$ Meet the Triangle Inequality Conditions?

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    2015
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The discussion centers on whether three positive real numbers \(x\), \(y\), and \(z\) can satisfy the triangle inequality conditions given the equation \(x^2 + 5y^2 + 4z^2 - 4xy - 4yz = 0\). Participants analyze the implications of the equation, exploring its geometric interpretation and potential solutions. Correct responses confirm that the numbers can indeed form the sides of a triangle, with justifications provided through algebraic manipulation and the properties of inequalities. The thread highlights the importance of understanding the relationship between the given equation and the triangle inequality. Overall, the conclusion is that the conditions are satisfied, allowing \(x\), \(y\), and \(z\) to represent the sides of a triangle.
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Here is this week's POTW:

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Three positive real numbers $x,\,y,\,z$ are such that $x^2+5y^2+4z^2-4xy-4yz=0$. Can $x,\,y,\,z$ form the sides of a triangle? Justify your answer.-----

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Congratulations to the following members for their correct solutions:

1. kaliprasad
2. Opalg
3. lfdahl

Solution from lfdahl:

The given equation:

$x^2+5y^2+4z^4-4xy-4yz = 0$ can be written as: $(x-2y)^2 + (y-2z)^2 = 0$

This equation holds iff $x = 2y \: \: \: \wedge \: \: \: y = 2z.$

The general solution in terms of an arbitrary choice of $z$ is:

$(x,y,z) = (4z,2z,z), \:\:\: z \in \mathbb{R}_+$

Using the three numbers as lengths in a triangle would violate the triangle inequality theorem (with $x$ being the longest edge):

$x \leq y+z \Rightarrow 4z \leq 2z+z=3z$

Conclusion:

The equation $x^2+5y^2+z^4-4xy-4yz = 0$ has an infinite set of triplet solutions $(x,y,z)$, none of which
can form a triangle.
 
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