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RMO UP 2014 Problem on triple simultaneous equation

Find all positive real numbers x , y ,z

\(2x-2y+\frac{1}{z}=\frac{1}{2014}\)

\(2z-2x+\frac{1}{y}=\frac{1}{2014}\)

\(2y-2z+\frac{1}{x}=\frac{1}{2014}\)

After solving it for a long time whatever I do I cannot form a equation with just 2 variables , so that I can proceed ahead.Please help. This was one of the two questions I was not able to solve in this years RMO.

Note by Sudhanshu Mishra
2 years, 3 months ago

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it is a very easy equation to solve only it need's is carefull observation first add all three equation you will get 1/x+1/y+1/z=3/2014 now transform all the equation by taking lcm and shifting denominator rhs for example first equation will become 2xz-2yz+1=z/2014 . now by adding all the three equation which we have transformed we will get x+y+z=3*2104. now we see that x,y,z, follows this equality (1/x+1/y+1/z)(x+y+z)=9 but we know that by AM-GM inequality that (x+y+z)(1/x+1/y+1/z) is greater than or equals to 9 and equality only holds when x=y=z this implies that x=y=z=2014 Onkar Tiwari · 2 years, 2 months ago

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I like this solution by @Joel Tan using Cauchy Schwarz. Calvin Lin Staff · 2 years, 3 months ago

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