Let #f(x) = (1+x^3)^(1/2)# and note that #f(0) = 1#.
Furthermore, #f'(x) = (3x^2)/(2sqrt(1+x^3))# which is always positive, so #f(x)# is increasing.
Therefore #f(0)=1# is the absolute minimum for #f# on #x >=0#.
And #f(x) > 1#
Now let #g(x) = (1+x^3)-(1+x^3)^(1/2)#.
Observe that #g(0) = 0#.
Now show that #g'(x) > 0# for #x >= 0#, so #0# is the minimum for #g(x)# on #[0,oo)#. (We'll need #x >= 0# to get #1/(2sqrt(1+x^3)) < 1#.)
The result follows.
Another approach
For #x >= 0#, we get #1+x^3 >= 1#.
Let #u = 1+x^3# and consider #u >= 1#.
For #1 <= u#, we also have #1 <= sqrtu# (because the square root function is an increasing function.)
Since, #1 <= sqrtu# we can multipl both sides by #sqrtu# to get
#sqrtu <= sqrtu^2 = u#.
Undoing the substitution, we get
#1 <= sqrt(1+x^3) <= 1+x^3# for #x >=0#.
