On the decreasing sequences I do not know how to prove that these sequences are decreasing:
$$
a_n=\frac{1}{\sqrt{n}(n+1)}-\frac{1}{\sqrt{n+1}(n+2)},
$$
$$
b_n=\frac{1}{n(\sqrt{n}+1)}-\frac{1}{(n+1)(\sqrt{n+1}+1)}.
$$
Thank you for all kind help and comments.
My attemption. I considered the following function
$$
f(x)=\frac{1}{\sqrt{x}(x+1)}-\frac{1}{\sqrt{x+1}(x+2)},
$$
and calculated its derivative. But its derivative is rather complicated
$$
\nabla f(x)=\frac{3x+1}{2\sqrt{x}x(x+1)^2}-\frac{3x+4}{2\sqrt{x+1}(x+1)(x+2)^2}
$$

Both sequences can be written as $a_n = f(n) - f(n+1)$ where $f$
is a (stricly) convex function.
Then
$$
 a_{n+1} - a_n = \bigl(f(n+1) - f(n+2)\bigr) - \bigl((f(n) - f(n+1)\bigr) = 2f(n+1)- f(n) - f(n+2) < 0
$$
because 
$$
f(n+1) < \frac12 \bigl(f(n) + f(n+2) \bigr)
$$
holds for a strictly convex function.
For your first sequence, $f(x)=\frac{1}{\sqrt{x}(x+1)}$
is convex because 
the second derivative
$$
f''(x) = \frac{5x^2 + 10 x + 3}{4 x^{5/2}(x+1)^3}
$$
is positive for $x > 0$.
For the second sequence, $f(x) = \frac{1}{x (\sqrt{x} + 1)}$
and
$$
f''(x) = \frac{15 x + 21 \sqrt x +8}{4 (x^{3/2} + x)^3} 
$$
is positive for $x > 0$.
(All derivatives calculated with Wolram Alpha :)
