at least $a\sqrt{n}$ solutions 
Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed. 

Consider a number  $0<a<2\pi$. How can we show that there is an infinite amount of integers $n$ such that 
$x^2 + y^2 + z^2=n$ 
has at least $a\sqrt{n}$ integer solutions?
I apologize for not editing in the boldened bit earlier. 
 A: Assuming that $a$ and $b$ are fixed and that $x^2+y^2+z^2 = n$ has $\le a \sqrt{n}+b$ integer solutions for all $n$, we get that the number of integer solutions of $x^2+y^2+z^2 < n$ is bounded above by
$$\sum_{k=0}^{n-1} (a \sqrt{k}+b) = bn+ an^{3/2}\sum_{k=0}^{n-1} \sqrt{\frac{k}n} \frac1n \le bn+ a n^{3/2} \int_0^1 \sqrt{x} \, dx = bn+\frac{2an^{3/2}}{3},$$
since the second sum is a lower Riemann sum for the integral of $\sqrt{x}$ over $[0,1]$.
Now since the volume of the ball with radius $\sqrt{n}$ is $\frac{4\pi n^{3/2}}3$, and the number of integer solutions of $x^2+y^2+z^2 <n$ has the same asymptotic growth, we get that $\frac{2a}3 \ge \frac{4\pi}3$, i.e., $a \ge 2\pi$.
Lastly, if there are only finitely many $n$ such that $x^2+y^2+z^2 = n$ has $\ge a\sqrt{n}$ integer solutions, it is straightforward that there exists $b$ such that $x^2+y^2+z^2 = n$ has $\le a\sqrt{n}+b$ integer solutions for all $n$.
A: The number of triples $(x,y,z)$ with $x^2+y^2+z^2\le n$ is roughly the volume of a sphere of radius $\sqrt n$, which is $(4/3)\pi n^{3/2}$. So there must exist $n$ with at least $(4/3)\pi\sqrt n$ solutions to $x^2+y^2+z^2=n$. That's not what you want, but it's a start, and maybe a refinement of this argument gets you where you want to go. 
EDIT: Let's try to refine the argument. Given a positive integer $m$, consider the triples with $m/2\le x^2+y^2+z^2\lt m$. The number of such triples is the volume of the region between the spheres of radius $\sqrt{m/2}$ and $\sqrt m$ (plus terms of lower order), and that's $${4\over3}\pi\Bigl(1-{\sqrt2\over4}\Bigr)m^{3/2}$$ Since we're talking about $m/2$ different integer values, (at least) one of those values must occur (at least) $(8/3)\pi(1-(\sqrt2/4))\sqrt m$ times. That's about $1.72\pi\sqrt m$, still not quite the $2\pi$ you want. Might be worth looking at a thinner shell. 
