proof that a function is decreasing I need to prove that $$\sqrt{3+x^{1/3}}+2\over x-1$$ is a decreasing function in $(1,\infty)$ with the definition of a decreasing function; the problem I have is that this expression is difficult so please I need your help
 A: First, thank you for the definition, David.
Now, back to the point on hand. Try looking at factors you can multiply this expression with to simplify it like let's say $\sqrt{3+x^{1/3}}-2$ so that it is sufficient to show that $$\frac{(\sqrt{3+x^{1/3}}+2)(\sqrt{3+x^{1/3}}-2)}{x-1}<\frac{(\sqrt{3+y^{1/3}}+2)(\sqrt{3+y^{1/3}}-2)}{y-1} \forall\; x>y.$$ Maybe you can simplify this further?
A: $$\sqrt{3+x^{1/3}}+2\over x-1$$
Convince yourself that this function is decreasing on $(1,\infty)$ if and only if (with the substitution $x^3$ instead of $x$) the function
$$\sqrt{3+x}+2\over x^3-1$$
is decreasing. Convince yourself again that this is decreasing on $(1,\infty)$ if and only if (with the substitution $x^2+4x+1$ instead of $x$)
$${\sqrt{x^2+4x+4}+2\over (x^2+4x+1)^3-1} = \frac{1}{x((x^2+4x+1)^2+(x^2+4x+1)+1)}$$
is decreasing. But hey, the final form is obviously decreasing on $(1,\infty)$ (We have used $x^2+4x+4 = (x+2)^2$ and $x^3 - 1 = (x-1)(x^2+x+1)$ for the last step).
A: You want to show that if x < y implies f(x) < f(y) ( if it is strictly decreasing -- nothing changes if we use $\le$ but it's more trouble to type.  
Comparing $\frac{\sqrt{3 + x^{1/3}} + 2}{x -1}$ to  $\frac{\sqrt{3 + y^{1/3}} + 2}{y -1}$ is the same as comparing $\frac{\sqrt{x^{1/3}}}{x}$ to  $\frac{\sqrt{y^{1/3}}}{y}$ (because the stuff I dropped out isn't relevant to which is larger).
This gives $\frac {x^{1/6}}{x}$ to be compared to $\frac {y^{1/6}}{y}$.  Since y > x > 1 we have $\frac{1}{x^{5/6}}$ > $\frac{1}{y^{5/6}}$
A: It oftentimes helps to make a change of variable that clears out some of the junk.  In this case, letting $u=\sqrt{3+x^{1/3}}$ turns the given expression into
$${u+2\over(u^2-3)^3-1}$$
Now note that 
$$\begin{align}
(u^2-3)^3-1&=(u^2-3)-1)((u^2-3)^2+(u^2-3)+1)\\
&=(u^2-4)(u^4-6u^2+9+u^2-2)\\
&=(u+2)(u-2)(u^4-5u^2+7)\\
\end{align}$$
This makes the expression in question
$${1\over(u-2)(u^4-5u^2+7)}$$
Note also that, as $x$ runs from $1$ to $\infty$, $u$ runs from $2$ to $\infty$ (and vice versa).  To show that the original expression is decreasing for $1\lt x\lt\infty$, therefore, it suffices to show that the two factors in the denomiator of the last expression are each increasing for $2\lt u\lt\infty$.  This is obvious for $(u-2)$, and it's an easy calculation for $(u^4-5u^2+7)$.
Added later:  Here is an alternative proof that replaces messy algebra with a chain of logical reasoning, relying on straightforward observations about increasing and decreasing functions:


*

*The sum of decreasing functions is decreasing.

*The sum of increasing functions is increasing.

*For functions that don't change sign, the reciprocal of a decreasing function is increasing, and the reciprocal of an increasing function is decreasing.

*For non-negative functions, any positive power of a decreasing function is decreasing, and likewise for increasing.


Starting from
$${\sqrt{1+x^{1/3}}+2\over x-1}={\sqrt{1+x^{1/3}}\over x-1}+{2\over x-1}$$
the $2/(x-1)$ is clearly decreasing on $(1,\infty)$, so it suffices to show that $(\sqrt{3+x^{1/3}})/(x-1)$ is decreasing.  Reciprocating and separating, it suffices to show that
$${x\over\sqrt{3+x^{1/3}}}-{1\over\sqrt{3+x^{1/3}}}$$
is increasing.  The $-1/\sqrt{3+x^{1/3}}$ is clearly increasing (it's negative with a denominator that's getting larger), so it suffices to show that the $x/\sqrt{3+x^{1/3}}$ is increasing.  Reciprocating, squaring, and separating, it suffices to show that
$${3+x^{1/3}\over x^2}={3\over x^2}+{1\over x^{5/3}}$$
is decreasing. But this is now obvious.
