(1) Proof 1: Given the function $f(x)=x+ \frac {a}{x}+b$, where $a$ and $b$ are constants and $a > 0$,
let's assume $0 0$,
thus $f(x_{1}) > f(x_{2})$,
therefore $f(x)$ is a monotonically decreasing function in the interval $(0, \sqrt {a}]$.

Proof 2: Given the function $f(x)=x+ \frac {a}{x}+b$, where $a$ and $b$ are constants and $a > 0$,
then $f′(x)=1- \frac {a}{x^{2}}= \frac {x^{2}-a}{x^{2}}$,
when $x\in(0, \sqrt {a}]$, $f′(x)\leqslant 0$ always holds,
thus $f(x)$ is a monotonically decreasing function in the interval $(0, \sqrt {a}]$.

(2) Given that the function $f(x)$ is monotonically increasing in the interval $[\sqrt {a},+\infty)$, and the maximum value of $f(x)$ in the interval $[1,2]$ is $5$, while the minimum value is $3$,
when $a\leqslant 1$, i.e., \begin{cases} 1+a+b=3 \\\\ 2+ \frac {1}{2}a+b=5\\end{cases}, solve for $a$: $a=-2$ (rejected);
when $1 < a\leqslant 2.25$, i.e., \begin{cases} 2 \sqrt {a}+b=3 \\\\ 2+ \frac {1}{2}a+b=5\\end{cases}, solve for $a$: $a=0$ (rejected), or $a=16$ (rejected);
when $2.25 < a < 4$, \begin{cases} 2 \sqrt {a}+b=3 \\\\ 1+a+b=5\\end{cases}, solve for $a$: $a=3+2 \sqrt {2}$ (rejected),
when $a\geqslant 4$, i.e., \begin{cases} 2+ \frac {1}{2}a+b=3 \\\\ 1+a+b=5\\end{cases}, solve for $a$: $a=6$;
therefore, $\boxed{a=6}$.