MHB Triangle Lengths: Can $a,b,c$ Form a Triangle?

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The equation given, $a^2 + 5b^2 + 4c^2 - 4ab - 4bc = 0$, can be analyzed to determine if the values of $a$, $b$, and $c$ can represent the sides of a triangle. By applying the triangle inequality, the conditions $a + b > c$, $a + c > b$, and $b + c > a$ must be satisfied. The equation can be rearranged and factored to explore the relationships between the variables. Ultimately, a solution must demonstrate that the values derived from the equation fulfill the triangle inequality criteria. Therefore, the conclusion hinges on whether valid positive values for $a$, $b$, and $c$ exist that satisfy both the equation and the triangle inequalities.
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Three positive real numbers $a,\,b$ and $c$ are such that $a^2+5b^2+4c^2-4ab-4bc=0$. Can $a,\,b$ and $c$ be the lengths of the sides of a triangle? Justify your answer.
 
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We have

$a^2+5b^2+4c^2-4ab-4bc=0$

Or $a^2-4ab + 4b^2 + b^2 - 4bc + 4c^2 = 0$

Or $(a-2b)^2 + (b- 2c)^2= 0$

Because a,b,c are real so each part is zero

This gives the solution a = 2b and b = 4c

Or a = 4c, b =2c and so $a > b+ c$

so the answer is no
 
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