my proof = add all three equations to get $ (1/x) + (1/y) + (1/z) = 3/2014 $ and taking lcm and cross multiplying and than adding we get $ x+y+z = 3(2014) $ now by AM-HM inequality we have $(x+y+z)((1/x) +(1/y)+(1/z)) = 3(2014)(1/2014)3 = 9 \geq 9 $ thus equality occurs. hence $x=y=z = 2014 $

aditya21 wrote: and taking lcm and cross multiplying and than adding we get $ x+y+z = 3(2014) $ What lcm? The numbers $x,y,z$ are real, not integer. You mean clearing denominators, then adding. Then there is the small issue that $x,y,z$ are not known to be positive, so applying AM-HM isn't quite justified. For example, denoting $x=2014a$, $y=2014b$, $z=2014c$, we get $a+b+c = 1/a+1/b+1/c = 3$, which also has other (not all positive) real solutions than $a=b=c=1$.

but mavpronevma AM-HM or cauchy schwarz inequality can be applied to reals i suppose it doesnt violate the condition of x,y,z being reals can you clarify? also by lcm i wanted to say like $ 2x-2y+(1/z) =1/2014$ can be wriiten as $ 2zx-2yz+1 = (z/2014)$ and similarly for others and than adding all 3 equations.

Yes, read my post above. There is for example the solution $a=\dfrac {-1-\sqrt{27/11}}{2}$, $b=\dfrac {-1+\sqrt{27/11}}{2}$, $c=4$. Cauchy-Schwarz can be applied for example in the form $(x^2+y^2+z^2)(r^2/x^2+s^2/y^2+t^2/z^2) \geq (r+s+t)^2$, even for some negatives among $x,y,z,r,s,t$, but not for $(x+y+z)(r^2/x+s^2/y+t^2/z) \geq (r+s+t)^2$. Just the relations for $x+y+z$ and $1/x+1/y+1/z$ are not enough; you need to also refer back to one of the original relations.

ohh thanks mavpronevma! so how do we solve this system of equations? also one of my friend wrote that since the equation is symmetric in $x,y,z$ than $x=y=z$ which is wrong according to me can you clarify?

Of course the symmetry argument is wrong; the system $x+y+z=6$, $xy+yz+zx=11$, $xyz=6$ is symmetric, but its solutions are $\{1,2,3\}$.

do you have a solution to this question mavropnevma?

Just do what you did to get \[\dfrac1x+\dfrac1y+\dfrac1z=\dfrac{3}{2014}\] and \[x+y+z=3(2014)\] Then let $xy+yz+zx=3k$ hence $xyz=2014k$. By Vieta's $x,y,z$ are roots of $f(n)=n^3-6042n^2+3kn-2014k$. Since $x,y,z$ are real, we know $\Delta \geq 0$ aka $-108 k^3+876138336 k^2-1776894406964928 k\geq 0$ which is just $(k-4056196)^2 k\leq 0$. Now the only positive solution is $k=4056196$ which gives the beloved solution $x=y=z=2014$ else $k\leq 0$ and you can use the cubic formula (although there will be negative solutions, if the problem wanted it solved in positive reals, it would've been much easier).

utkarshgupta wrote: Find all real $x,y,z $ such that $2x-2y+(1/z) = (1/2014)$ , $ 2y-2z +(1/x) = (1/2014) $ and $ 2z-2x+ (1/y) = (1/2014) $ the question asked to find positive reals so $x=y=z=2014$ is the answer

That was also my guess ... judging by the degree of difficulty of the other problems, it must have had this extra, simplifying condition.

FWIW : Finding positive real solution(s) is easy but finding all (even not restricted to real) solutions is not hard either. (Assuming solving a cubic equation is not hard ..specially if numerical (approximate) solution is sufficient) $2x-2y+(1/z) = (1/2014)$ , $ 2y-2z +(1/x) = (1/2014) $ and $ 2z-2x+ (1/y) = (1/2014) $ As pointed out, adding all we get $1/x+1/y+1/z=3/2014$ and writing the three as $2z(x-y)+1=z/2014$ ...etc , and add them together we have: $x/2014+y/2014+z/2014=3$ Again for simplicity (or general solution for any k) let $k=2014$, and $x=ka,y=kb,z=kc$ We have $a+b+c=3$ (eq 1) $1/a+1/b+1/c=3$ eq(2) and of course, $2k^2 c(a-b)+1=c$ (eq 3) We can simplify a bit by assuming $a=1+p,b=1+q,c=1+r$ so we get $p+q+r=0$ And $1/(1+p)+1/(1+q) = 3 - 1/(1+r)$ $\frac {2+p+q}{1+p+q+pq} = \frac {3+3r -1} {1+r}$ Substituting $p+q=-r$ to get this equation in pq and r .. we have $\frac {2-r}{1-r+pq} = \frac {2+3r}{1+r}$ which gives us $pq = \frac {2r^2} {2+3r}$ We have the values of p+q (=-r) and pq (in terms of r) and eq 3 gives the value of (p-q) Eq 3 now is $2k^2(1+r)(p-q)=c-1=r$ which gives us $p-q = (1/2k^2)(r/(1+r))$ so $(p+q)^2 = (p-q)^2+4pq$ now yields a nice equation in $r$ : $r^2 = \frac {(1/4k^4)r^2}{(1+r)^2} + \frac {8r^2} {2+3r}$ $r^2=0 $ is obviously a solution, other solutions are given by: $1 = \frac {(1/4k^4)}{(1+r)^2} + \frac {8} {2+3r}$ This simplify nicely to $12k^4 r^3 -(3+36k^4)r-(2+24k^4)$ Or $12k^4 r^3 -3(12k^4+1)r-2(12k^4+1)$ Which again comes out neatly as, $ r^3 - 3nr -2n$ where $n = 1+1/(12*2014^4)$ (Please note that n is very close to 1) The values (since $n$ is pretty close to $1$ ) of $r$ is close to -1,+1,2 (Because when $n=1$ reduces above to $r^3-3r-2 = (r+1)^2(r-2)=0$ This gives value of $c$ close to $0,0,3$ For the case of 2014, we have, The additional solutions for values $c$ are approximately $-.00000004..,+.00000004..,3.000000000000004$ etc So the solutions for $x,y,z$ are: $(2014,2014,2014)$ and close to (0,0,3*2014) More accurately $v$ of the order of $10^{-11}$ and $u$ is of the order of $10^{-4}$ and $(x,y,z)$ are $(-u,(u-v), (3*2014+v)$ where, $v << u <<1 $ (Cyclic combinations)

Sorry for reviving this post but I have found out another solution to this question and I want to know whether it is correct or it has some flaw W.L.O.G assume $x \geq y \geq z$, then $\frac{1}{z} \geq \frac{1}{y} => 2x-2y+\frac{1}{z} \geq 2z-2x+\frac{1}{y}$, equality iff $x=y=z$. As equality is given in the question, we get our only solution as $x=y=z=2014$.