×

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
3 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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

- 3 years, 2 months ago

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

Staff - 3 years, 3 months ago