Prove conjecture $a_{n+1}>a_{n}$ if $a_{n+1}=a+\frac{n}{a_{n}}$ Let sequence $\{a_{n}\}$ such $a_{1}=a>0$,and 
$$a_{n+1}=a+\dfrac{n}{a_{n}}$$
I used the software to find this following conjecture :
if $n>\dfrac{4}{a^3}$,we have 
$$a_{n+1}>a_{n}$$

Here is an answer which proves that the conjecture holds for $a > 1$. For the general case I couldn't show it but I present some ideas. 
The proof follows by induction. 
We want $a_{n+1} > a_n$. So we need (see the calculations by Han de Bruijn):
$$
a_n <  \frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n} = U(n)
$$
So this constitutes an  upper bound $U(n)$ for all $a_n$. Since this must hold for all $a_n$, we can ask for a condition for this to hold for  $a_{n+1}$. We get 
$$
a + \frac{n}{a_n} = a_{n+1} <  \frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n+1} = U(n+1)
$$
which gives (SOME MORE STEPS ADDED HERE by request)
$$
a_{n} > \frac{n}{- {a} + U(n+1) } = \frac{n}{n+1} U(n+1)
$$
The latter equality can be verified by clearing denominators:
$$
n+ 1 = {- {a} U(n+1) } + (U(n+1))^2
$$
inserting $U(n+1)$
$$
n+ 1 = - {a} \left[\frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n+1} \right] + \left[\frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n+1} \right]^2
$$
and expanding the brackets:
$$
n+ 1 = - \frac{a^2}{2} - a \sqrt{\left(\frac{a}{2}\right)^2+n+1}  + \frac{a^2}{4} + a \sqrt{\left(\frac{a}{2}\right)^2+n+1} + \left(\frac{a}{2}\right)^2+n+1
$$
which clearly is an identity. (END EDIT)
Summarizing, we have
$$
a_n >  \frac{n}{n+1} \left[\frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n+1} \right] = \frac{n}{n+1} U(n+1) = L(n)
$$
So this constitutes a lower bound $L(n)$ for all $a_n$. By induction, if for some $N$, $L(N) < a_N < U(N)$, then these lower and upper bounds are necessary conditions for all further $n > N$ in order that  $a_{n+1} > a_n$. 
Now we give a (further) sufficient condition for that to hold. The idea is to show that for any $a_n$ with   $L(n) < a_n < U(n)$, also   $L(n+1) < a_{n+1} < U(n+1)$  holds.  Since         $a_{n+1} = a + \frac{n}{a_n}$, we have that 1) the highest  $a_{n+1}$ will be obtained from the lowest $a_{n} = L(n)$ and 2) 
the lowest  $a_{n+1}$ will be obtained from the highest $a_{n} = U(n)$.
So we need for 1) 
$$
a + \frac{n}{L(n)} \leq U(n+1)
$$
and for 2) 
$$
a + \frac{n}{U(n)} \geq L(n+1)
$$
Establishing this,  we have, by induction, that all $a_n$ will always stay within the limits $L(n) < a_n < U(n)$ and hence we have proved the OP's claim.
For 1): Inserting $L(n)$ and $U(n+1)$ gives 
$$
a + \frac{n}{\frac{n}{n+1} U(n+1)} \leq U(n+1)
$$
Solving this for $U(n+1)$ gives 
$$
U(n+1) \geq \frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n+1}
$$
which holds by the definition of $U(n)$. 
For 2): Inserting $L(n+1)$ and $U(n)$ gives 
$$
a + \frac{n}{U(n)} \geq \frac{n+1}{n+2} U(n+2) 
$$
The LHS equals $U(n)$ (clear by inserting it). Then we need
$$
{U(n)} \geq \frac{n+1}{n+2} U(n+2) 
$$
Inserting $U(n)$ and doing the algebra gives 
$$
a \sqrt{\left(\frac{a}{2}\right)^2+n} > 1 - \frac{a^2}{2}
$$
Clearly, for $a > 1$ this holds. For $a < 1$ we can square and get 
$$
a^2 \left(\frac{a}{2}\right)^2+ n a^2 >  \frac{a^4}{4} + 1 - a^2
$$
or 
$$
n > \frac{1}{a^2} -1
$$
Let's first look at the case $a > 1$. This gives for $n=1$ (start of induction) that $ \frac12 \left[\frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+2} \right] = L(1) < a_1 < U(1) = \frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+1}$
Since $a_1 = a$ by the task description, we must have that 
$ \frac12 \left[\frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+2} \right] < a < \frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+1}$
The right inequality is obvious and the left inequality holds for $a > 1$. This establishes my first claim that the OP's conjecture holds for $a > 1$.
As for $a <1$, in this case we have that the initial condition $a_1=a$ violates the lower bound $L(1)$. Further, our sufficient (conservative) condition will only be satisfied for $
n > \frac{1}{a^2} -1
$. 
If we require, as in the original setting, $n > \frac{4}{a^3}$, then for $a <1$, the condition  $
n > \frac{1}{a^2} -1
$ will be satisfied anyway. So the program of proof could be:
show that, if we start with $a_1 = a$, and take $N = \frac{4}{a^3}$ many iteration steps $a_{n+1} = a + n/a_n$, then  $L(N) < a_N < U(N)$. From this the general proof would follow.
