<s>[INST] Find the value of $a$ 
Find $a$ for which $f(x) = \left(\frac{\sqrt{a+4}}{1-a} -1\right)x^5-3x+\ln5 \;$ decreases for all $x$ with $a\neq 1$ and $a\geq -4$.   

My try:
For $f(x)$ to decrease
$$5x^4\left(\frac{\sqrt{a+4}}{1-a} -1\right) -3 <0\implies x^4\left(\frac{\sqrt{a+4}}{1-a} -1\right) < 3/5 $$ 
How can I proceed further?
 [/INST] HINT
We have
$$f(x) = \left(\frac{\sqrt{a+4}}{1-a} -1\right)x^5-3x+\ln5 \implies f'(x) = 5\left(\frac{\sqrt{a+4}}{1-a} -1\right)x^4-3\le 0$$
then we need
$$\left(\frac{\sqrt{a+4}}{1-a} -1\right)x^4- \frac35 \le 0 $$
that is true for all $x$ when
$$\frac{\sqrt{a+4}}{1-a} -1\le0 \iff \frac{\sqrt{a+4}+a-1}{1-a}\le0$$
</s>