Try

1) We want to show that

1/x+1/y+1/z >= (3a+1)(1/(x+a) + 1/(y+a) + 1/(z+a)) = f(a), where a >= 0

Which is to show

（1-3x)/(x(x+a)) + (1-3y)/(y(y+a)) + (1-3z)/(z(z+a))
This looks ugly but in fact not:

Assume x >= y >= z, then u=1/（x(x+a))
So by the rearrangement inequality, xu + yv + zw
2) Let a = 1/2 in the above inequality:

1/x+1/y+1/z >= (3/2+1)(1/(x+1/2) + 1/(y+1/2) + 1/(z+1/2)) >= 5(1/（x+y+1) +...), by let a=x+1/2, b=y+1/2, c=z+1/2, and using inequality of the form 1/a + 1/b >= 4/(a+b).