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I have an equation $(a-2b)^2 + (b-2c)^2 = 0$ where, $a$, $b$ and $c$ are positive real numbers. What will be $a:b:c$?

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  • $\begingroup$ What simple inequalities does $x^2$ obey $\forall x$? $\endgroup$
    – A Simmons
    Dec 7, 2015 at 17:34
  • $\begingroup$ it will always be positive and for x=0, x^2 =0. okay does it mean a = 2b and b = 2c.? thank you I understood this... $\endgroup$
    – Rudra
    Dec 7, 2015 at 17:36

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For any real number $x$ we have $x^2\ge 0$, thus it follows that $a-2b=b-2c=0.$ Can you take it from here?

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  • $\begingroup$ Yes, Thank you Jack. $\endgroup$
    – Rudra
    Dec 7, 2015 at 17:38

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