Is there either an explicit formula or a numerical method for solving an equation in the form
$$ \sum_{n=1}^{N}{\left. \sqrt{x+{a_n}}\right.}=\mathit{ b} $$
over the real numbers? (a method that gives complex solutions would also be useful, so I would select the real ones)
If that's not possible for some reason, then giving an estimate solution to the inequality form would also be useful for me:
$$ \sum_{n=1}^{N}{\left. \sqrt{x+{a_n}}\right.} ≤ \mathit{ b} $$
I tried to use some inequalities (from here) to simplify it, without success. (I'm no mathematician.) And I'm not even sure how close the estimate would be.
I need this to get a step further in the following problem: Maximizing sum of polynomials of degree 3
Many thanks!
Assuming that all $a_n$'s are real and positive, without loss of generality, less us consider $0 \leq a_1 \leq a_2 \leq \cdots \leq a_{n-1} \leq a_n$ (if this is not the case, a simple change of variable $y=x+a_1$ will make it).
So,
$$N \sqrt{x+{a_1}}\leq \sum_{n=1}^{N} \sqrt{x+{a_n}} \leq N \sqrt{x+{a_N}}$$ $$x_{min}=\frac{b^2}{N^2}- a_N \leq x \leq \frac{b^2}{N^2}- a_1=x_{max}$$
So we have clear bounds for $x$ and Newton method would converge quite fast.
Since we face an increasing function with negative second derivative, if want to avoid any overshoot of the solution, according to Darboux theorem, let us start at the left bound. If this is not a concern, start at the midpoint of the interval.
For illustration purposes, consider $N=123$, $a_n=p_n$ and $b=12345$. So, we have $$9396.3 \leq x \leq 10042.3$$
The iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 9396.3045806067816776 \\ 1 & 9760.0603412690434146 \\ 2 & 9763.4700434412188284 \\ 3 & 9763.4703322248426733 \\ 4 & 9763.4703322248447441 \end{array} \right)$$
Edit
To have a better approximation of the solution, we could look for an equivalent solution writing $$\sum_{n=1}^{N}\int_{x_{min}}^{x_{max}} \sqrt{x+{a_n}} \,dx=N \int_{x_{min}}^{x_{max}} \sqrt{x+{\hat a}} \,dx$$ leading to $$\sum_{n=1}^{N} \big[(x_{max}+a_n)^{\frac32}-(x_{min}+a_n)^{\frac32}\big]=N\big[(x_{max}+\hat a)^{\frac32}-(x_{min}+\hat a)^{\frac32}\big]$$ Solving for $\hat a$ would give $$x\approx\frac{b^2}{N^2}-\hat a$$ and I suppose that one single Newton iteration would suffice.
For the worked example, this would give $\hat a=309.83057$ and $x=9763.47400$.
Update
I cannot prove it but a large number of random tests seem to show that an appropriate lower bound could be $$x'_{min}=\frac{b^2}{N^2}-\frac{\sum_{n=1}^N a_n } N$$ and that, for this value $y<0$. For the worked example, $x'_{min}=9762.43$.