The expansion of (a+b)^u is a^u+ua^(u-1)b+...+ub^(u-1)a+b^u.
If we subtract a^u+b^u from this expansion we're left with all the other positive terms (assuming a,b≠0). Let the sum of these terms be S.
So (a+b)^u=a^u+b^u+S and S>0. (a+b)^u-(a^u+b^u)>0. Therefore a^u+b^u<(a+b)^u.