<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] This holds true for every $x$ only if $$\frac{\sqrt{a+4}}{1-a} -1\le 0$$or $${\sqrt{a+4}\over 1-a}\le 1$$Surely $a>1$ is one region of answer. For $a<1$ we have $$\sqrt {a+4}\le 1-a\to a+4\le a^2-2a+1\to a^2-3a-3\ge 0\to\\a\ge {3+\sqrt {21}\over 2}\\a\le {3-\sqrt {21}\over 2}$$therefore the region of the answer is $$[-4, {3-\sqrt {21}\over 2}]\cup (1,\infty)$$
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