How to estimate upper bound of $|\frac{3h^2k}{6(h^2+k^2)^{3/2}}|$ to $\frac{3}{6}?$

Question

This is in my textbook as part of a proof for Taylor polynomial in multivariable calculus.

I understand that you can estimate the following expression by removing $k^2$ from the denominator:

$$|\frac{h^3}{6(h^2+k^2)^{3/2}}| \leq |\frac{h^3}{6(h^2)^{3/2}}| = \frac{1}{6}$$

But how exactly could you estimate $|\frac{3h^2k}{6(h^2+k^2)^{3/2}}| \leq \frac{3}{6}$ ?

Answer 1

Because by AM-GM $$\left|\frac{h^2k}{\sqrt{(h^2+k^2)^3}}\right|=\left|\frac{h^2k}{\sqrt{(2\cdot\frac{h^2}{2}+k^2)^3}}\right|\leq\left|\frac{h^2k}{\sqrt{\left(3\sqrt[3]{\left(\frac{h^2}{2}\right)^2k^2}\right)^3}}\right|=\frac{2}{3\sqrt3}<1.$$ Id est, $$\left|\frac{3h^2k}{6\sqrt{(h^2+k^2)^3}}\right|<\frac{3}{6}=\frac{1}{2}.$$

Answer 2

$|k|=\sqrt{k^2}\le \sqrt{k^2+h^2};$

$\dfrac{3h^2|k|}{6(k^2+h^2)^{3/2}}\le$

$\dfrac{3h^2(k^2+h^2)^{1/2}}{6(k^2+h^2)^{3/2}}=$

$\dfrac{3h^2}{6(k^2+h^2)} \le 1/2$;

Note: $\dfrac{h^2}{h^2+k^2} \le 1$.

Answer 3

With only high-school tools:

Due to the absolute value, we may as well suppose $h,k >0$. Factoring $k^2$ in the denominator, we get: \begin{align} \Biggl|\frac{3h^2k}{6(h^2+k^2)^{3/2}}\Biggr|= \frac{3h^2k}{6k^3\biggl(\cfrac{h^2}{k^2}+1\biggr)^{3/2}}=\frac36\,\frac{\cfrac{h^2}{k^2}}{\biggl(\cfrac{h^2}{k^2}+1\biggr)^{\!\!3/2}}, \end{align} and the last fraction $\le 1$ because $$\frac{h^2}{k^2}<\frac{h^2}{k^2}+1,\qquad \biggl(\frac{h^2}{k^2}+1\biggr)^{\!\!3/2}>\frac{h^2}{k^2}+1. $$