How to find the partial derivative of this complicated function? Find the partial derivative with respect to a (treat other variables as constants):
$\displaystyle f(a,b,c)=\frac {a}{\sqrt{a^2+8bc}}-\frac {a^r}{a^r+b^r+c^r}$
The article I'm reading says it should be:
$\displaystyle \frac {\sqrt{a^2+8bc}-\dfrac{a^2}{\sqrt{a^2+8bc}}}{a^2+8bc}-\frac{ra^{r-1}(a^r+b^r+c^r)-a^r \cdot ra^{r-1}}{(a^r+b^r+c^r)^2}$
but I don't know how to get there.
Thanks in advance!
 A: Form the partial derivatives quotient rule, we know:
$$\displaystyle \frac{\partial}{\partial a} \left(\frac{g}{h}\right) = \frac{g' h - g h'}{h^2}$$
Given the function:
$$\displaystyle f(a,b,c)=\frac {a}{\sqrt{a^2+8bc}}-\frac {a^r}{a^r+b^r+c^r}$$
Lets do each term separately, so we have:
$$\frac{(1) \sqrt{a^2+8bc}-(a)(\frac{1}{2})(a^2+8bc)^{-1/2}(2a)}{a^2 + 8 bc} = \frac {\sqrt{a^2+8bc}-\dfrac{a^2}{\sqrt{a^2+8bc}}}{a^2+8bc}$$
Of course, we can reduce this further to:
$$\dfrac{8 b c}{(a^2+8 b c)^{3/2}}$$
For the second one we have:
$$\displaystyle \frac{r a^{r-1}(a^r + b^r + c^r) - a^r(r a^{r-1})}{(a^r + b^r + c^r)^2}$$
Of course we can reduce this to:
$$\dfrac{r a^{r-1} (b^r+c^r)}{(a^r+b^r+c^r)^2}$$
The final result would be:
$$\displaystyle \frac{\partial}{\partial a} f(a, b, c,) = \dfrac{8 b c}{(a^2+8 b c)^{3/2}} - \dfrac{r a^{r-1} (b^r+c^r)}{(a^r+b^r+c^r)^2}$$
A: $$\dfrac{a}{\sqrt{a^2+8bc}}\ge\dfrac{a^r}{a^r+b^r+c^r}\cdots \cdots (1)$$
I guess you do this show that
$$\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ca}}+\dfrac{c}{\sqrt{c^2+8ab}}\ge 1$$
so you can Substituting $b=c=1$ in (1)
we obtain the following condition on $r:f(a)\ge 0$ for all $a>0$,wher
$$f(a)=a(a^r+2)-a^r\sqrt{a^2+8}$$
Note that $f(1)=0,$
use Lemma 3:https://www.awesomemath.org/wp-content/uploads/reflections/2009_1/MR_1_2009_article1.
so $$f'(a)=(r+1)a^r+2-ra^{r-1}\sqrt{a^2+8}-a^r\cdot\dfrac{a}{\sqrt{a^2+8}}$$
so $$f'(1)=0\Longrightarrow  r=\dfrac{4}{3}$$
