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

[anemone]

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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[anemone]

Congratulations to the following members for their correct solutions:

1. kaliprasad
2. Opalg
3. lfdahl

Solution from lfdahl:

Spoiler

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.