Equal in proportion

if we sum above three equation, we can get

$\frac 1 x + \frac 1 y + \frac 1 z = \frac 3 {2014}$ .

And If we calculate $(1 equation) \times z + (2 equation) \times x + (3 equation) \times y$ ,

we can also get $x + y + z = 3 \times 2014$ .

with 'Arithmetic Mean' $\ge$ 'Harmonic mean Inequality' $(x_1 + x_2 + x_3 + ... + x_n) \times (\frac{1} {x_1} + \frac{1} {x_2} + \frac{1} {x_3} + ... + \frac{1} {x_n}) \ge n^2$

with equality if and only if $x_1=x_2=\cdots=x_n$ .

$(\frac 1 x + \frac 1 y + \frac 1 z) \times (x + y + z ) = 3^2$ so $x = y = z$ than $x + yz - zx = 2014$

add all the three equations. you'll end up getting (1/x) + (1/y) + (1/z) = (3/2014) .. let us assume that x=y=z=2014......... and then calculate x+ yz - zx

Any permutation of x,y,z leaves the equations unaffected. If any 2 of x,y,z are different, some permutations will affect the 3 numbers x,y and z. This contradicts the fact that the equations are unaffected. So x=y=z. However, this means that x=y=z=2014.

Just observe that $x=y=z=2014$ and so get the needed expression.