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

Dec 15, 2016

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

#### Explanation:

Let $f \left(x\right) = {\left(1 + {x}^{3}\right)}^{\frac{1}{2}}$ and note that $f \left(0\right) = 1$.

Furthermore, $f ' \left(x\right) = \frac{3 {x}^{2}}{2 \sqrt{1 + {x}^{3}}}$ which is always positive, so $f \left(x\right)$ is increasing.

Therefore $f \left(0\right) = 1$ is the absolute minimum for $f$ on $x \ge 0$.

And $f \left(x\right) > 1$

Now let $g \left(x\right) = \left(1 + {x}^{3}\right) - {\left(1 + {x}^{3}\right)}^{\frac{1}{2}}$.

Observe that $g \left(0\right) = 0$.

Now show that $g ' \left(x\right) > 0$ for $x \ge 0$, so $0$ is the minimum for $g \left(x\right)$ on $\left[0 , \infty\right)$. (We'll need $x \ge 0$ to get $\frac{1}{2 \sqrt{1 + {x}^{3}}} < 1$.)

The result follows.

Another approach

For $x \ge 0$, we get $1 + {x}^{3} \ge 1$.

Let $u = 1 + {x}^{3}$ and consider $u \ge 1$.

For $1 \le u$, we also have $1 \le \sqrt{u}$ (because the square root function is an increasing function.)

Since, $1 \le \sqrt{u}$ we can multipl both sides by $\sqrt{u}$ to get

$\sqrt{u} \le {\sqrt{u}}^{2} = u$.

Undoing the substitution, we get

$1 \le \sqrt{1 + {x}^{3}} \le 1 + {x}^{3}$ for $x \ge 0$.

Dec 15, 2016

See below.

#### Explanation:

For $u \ge 1$ we have $u \le {u}^{2}$ now if $g \left(\cdot\right)$ is a monotonically strictly increasing function we have

$g \left(u\right) \le g \left({u}^{2}\right)$

now making $g \left(\cdot\right) = \sqrt{\cdot}$ we have

$\sqrt{u} \le u$. Finally proposing

$u = 1 + {x}^{3} \ge 1$ we have

$1 \le \sqrt{1 + {x}^{3}} \le 1 + {x}^{3}$