easy but interesting.

but ur question is incomplete.....
you should hv added that x,y and z are positive real numbers...
and not only real numbers...
as A.M-G.M inequality holds only for positive real numbers.

I have a better solution. In my solution I only need x,y,z to be Reals only.

From 1st eqn.
z=1-x-y

Putting the value of z in 2nd eqn.
2xy-z²=1
→2xy-(1-x-y)²=1
which is equivalent to
1+x²+y²-2x-2y=-1
→(x-1)²+(y-1)²=0
→x=1 and y=1
so, z=-1
(x,y,z)=(1,1,-1)

yea that's a good solution Arnab..!!

@rahulmishra
yeah thanks you pointed it out.
here is another one.
x+y = 1-z
so, x^2 + y^2 + 2xy = 1 + z^2 - 2z
also 2xy = 1 + z^2
subtracting the latter equation form the former yields
x^2 + y^2 = -2z
subtracting - 2xy form both the sides.
(x-y)^2 = -2z - 2xy
or, (x-y)^2 = -2z - 1 - z^2 ( since 2xy = 1 + z^2)
or, (x-y)^2 + (z+1)^2 = 0
so, x = y and z = -1.

\hspace{-16}\mathbf{2xy=1+z^2}$\\\\ and $\mathbf{x+y+z=1\Leftrightarrow (x+y)=1-z}$\\\\ $\mathbf{*x^2+y^2\geq 2xy}$\\\\ $\mathbf{x^2+y^2+2xy\geq 4xy}$\\\\ $\mathbf{(x+y)^2\geq 2.(2xy)}$\\\\ $\mathbf{(1-z)^2\geq 2+2z^2\Leftrightarrow 1+z^2-2z\geq 2+2z^2}$\\\\ $\mathbf{z^2+2z+1\leq 0\Leftrightarrow (z+1)^2\leq 0}$\\\\ Means $\mathbf{1+z=0\Leftrightarrow z=-1}$\\\\ $\mathbf{(xy=1)}$ and $\mathbf{x+y=2}$\\\\ Means $\mathbf{x=y=1}$\\\\ So $\mathbf{x=y=1\;,z=-1}$\\\\ Given by hsbhatt Sir.......