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.

#Algebra

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Comments

I like this solution by @Joel Tan using Cauchy Schwarz.

Calvin Lin Staff - 6 years, 6 months ago

A long-awaited thank you.

Sudhanshu Mishra - 1 year ago

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 - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$