Prove that (xyz)^1/3[1/x +1/y +1/z] > 3 where x,y, z are +ve real nos.

AM >or= GM

AM of 1/x , 1/y , 1/z  =  (1/x + 1/y + 1/z) / 3

their GM = (1/xyz)^(1/3)

rearrange to get given expression

GM >= HM

(xyz)^1/3 >= 3/(1/x + 1/y + 1/z)

(xyz)^1/3 (1/x + 1/y + 1/z) >= 3 .............[proved]

G.P.>H.P.

(xyz)1/3>3/(1/x+1/y+1/z)

Hence proved.

you have mentioned that all are positive real nos...so this gives me a thought about AM greater than or equal to GM.

Consider 1/x, 1/y ,1/z as 3 real nos.... their am is (1/x+1/y+1/z)/ 3 and their GM is  cube root of (1/xyz) and since 

AM is greater than or equal to Gm 

u get their desired result.