Show that #1<=(1+x^3)^(1/2)<=1+x^3# for #x>=0# ?

2 Answers
Dec 15, 2016

Answer:

There are probably more elegant solutions, but here are two.

Explanation:

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#.

Dec 15, 2016

Answer:

See below.

Explanation:

For #u ge 1# we have #u le u^2# now if #g(cdot)# is a monotonically strictly increasing function we have

#g(u) le g(u^2)#

now making #g(cdot)= sqrt(cdot)# we have

#sqrt(u) le u#. Finally proposing

#u = 1+x^3ge 1# we have

#1 le sqrt(1+x^3) le 1 + x^3#