Prove that $\frac{1}{7\sqrt{2}} \leq \int_{0}^1 \frac{x^6}{\sqrt{1+x^2}}dx \leq \frac{1}{7}$ 
Prove that $\displaystyle \dfrac{1}{7\sqrt{2}} \leq \int_{0}^1 \dfrac{x^6}{\sqrt{1+x^2}}dx \leq \dfrac{1}{7}$.

My book says let $f(x) = \dfrac{1}{\sqrt{1+x^2}}$ and $g(x) = x^6$. Then $\displaystyle \int_0^1 \dfrac{x^6}{\sqrt{1+x^2}}dx = \dfrac{1}{\sqrt{1+\xi^2}} \int_{0}^1 x^6 dx$ where $0 \leq \xi \leq 1$. Thus $$\dfrac{1}{7\sqrt{2}} \leq \dfrac{1}{\sqrt{1+\xi^2}} \leq \dfrac{1}{7}.$$
I don't understand how they are getting all these results. The problem did say to use other results in the book, but I didn't see any that related. For reference it is in chapter 13 of Michael Spivak's calculus book.
 A: We have $$\frac{1}{\sqrt 2} \le \frac{1}{\sqrt{1+x^2}} \le 1$$ for all $x \in [0,1]$ since $0 \le x^2 \le 1$. Indeed, then $1 \le 1+x^2 \le 2$ and so $$\frac{1}{2} \le \frac{1}{1+x^2} \le 1.$$ Thus $$\frac{1}{\sqrt 2} \le \frac{1}{\sqrt{1+x^2}} \le 1.$$
Then just multiply by $x^6$ and integrate.
A: Achtung: overkill. $g(x)=\frac{1}{\sqrt{1+x}}$ is a convex function on $[0,1]$ (since its logarithm is convex) bounded between $1$ and $\frac{1}{\sqrt{2}}$. Since $g'(1)=-\frac{1}{4\sqrt{2}}$,
$$\forall x\in(0,1),\qquad \frac{(1-x)}{4\sqrt{2}}+\frac{1}{\sqrt{2}}\leq  g(x)\leq (1-x)+\frac{x}{\sqrt{2}}\tag{1}$$
holds by convexity. Now:
$$ I=\int_{0}^{1}\frac{x^6}{\sqrt{1+x^2}}\,dx =\frac{1}{2}\int_{0}^{1} x^{5/2}g(x)\,dx \tag{2}$$
can be bounded by using $(1)$:

$$ \color{red}{\frac{1}{10}}<\frac{19}{126\sqrt{2}} < \color{red}{I} < \frac{4+7\sqrt{2}}{126}<\color{red}{\frac{1}{9}}. \tag{3}$$

If we use integration by parts and the Cauchy-Schwarz inequality we also get the interesting lower bound:
$$ I = \sqrt{2}-\int_{0}^{1}5x^4\sqrt{1+x^2}\,dx \geq \sqrt{2}-\sqrt{\frac{12}{7}}.\tag{4} $$
