<s>[INST] Is this function decreasing? I have to prove that the following function is decreasing when $x>0$
$$f(x)=\frac{(x+1)^2}{\sqrt{x^2+2x}}-(x+1)^2\arcsin\Big(\frac{1}{x+1}\Big)$$
but calculating its derivative I don't succed in proving this result.
 [/INST] $f'(x) = \dfrac{3(x+1)-\dfrac{(x+1)^3}{(x+2)x}}{\sqrt{x(x+2)}}-2(x+1)\arcsin(\dfrac{1}{1+x})$
Since $1+x>0$,thus we just prove
$\dfrac{3-\dfrac{(1+x)^2}{(x+2)x}}{\sqrt{x(x+2}}-2\arcsin(\dfrac{1}{x+1})\le 0$
Since $\arcsin(\dfrac{1}{1+x})\ge \dfrac{1}{1+x}$, thus 
we just prove
$\dfrac{3-\dfrac{(1+x)^2}{(x+2)x}}{\sqrt{x(x+2)}}-\dfrac{2}{x+1}\le 0$
which is equivalent to
$3-\dfrac{(x+1)^2}{x(x+2)}-\dfrac{2\sqrt{x(x+2)}}{x+1}\le 0$
take $u = \dfrac{1+x}{\sqrt{x(x+2)}}$, then
we just have to show
$3-u^2-\dfrac{2}{u}\le 0$.
AM-GM Inequality
$u^2+\dfrac{1}{u}+\dfrac{1}{u}\ge 3\sqrt[3]{1}=3$
And we know $u\neq 1$, thus the equality cannot be achieved.
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