(x+y)^z+(y+z)^x+(z+x)^y > 2.

Without loss of generalization, we can assume that  x > y > z.

So, (x + y) > 2y > 2z.

And (x + z) > 2z.

And (y + z) > 2z.

Therefore the sum at left (let's call it S) is greater than :

(2z)^z + (2z)^z + (2z)^z = 3 x (2z)^z = 3 x 2^z x z^z

S > 3 x 2^z x z^z

When z tends to zero :

lim (S) = 3 x lim (2^z) x lim z^z, each limit when z-->0.

Since lim (2^z) = 1, and lim (z^z) is also = 1, then :

lim (S) = 3, when z--> 0.

Therefore, S > 2, as asked, and indeed >= 3.