System of equations with radicals: $2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}$ and $2\sqrt[4]{\frac{y^4}{3}+4} = 1+\sqrt{\frac{3}{2}x^2}$ 
Solve the system of equations (in $\mathbb R$):
  $$\begin{matrix}
2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}
\\
2\sqrt[4]{\frac{y^4}{3}+4}
=
1+\sqrt{\frac{3}{2}x^2}
\end{matrix}.$$

This is from an older question, which was closed and deleted because of lack of context. (Here is link for users who can see deleted questions.) I will post my solution below - so I hope this time the question will not be closed for the lack of effort.
I found the system not too easy and somewhat interesting. (Of course I might have missed some straightforward way to the solution.) I'd be glad to see some other methods to solve it.

Without loss of generality, we can assume that $x,y\ge0$. (Notice that all exponents are even. We can then add signs to get the remaining solutions.)
So our equations are changed to
$$2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}}y\tag{1}$$
$$2\sqrt[4]{\frac{y^4}{3}+4}=1+\sqrt{\frac{3}{2}}x\tag{2}$$
If we subtract the two equations, we get (1)-(2):
$$2\left(\sqrt[4]{\frac{x^4}{3}+4}-\sqrt[4]{\frac{y^4}{3}+4}\right)=y-x.\tag{3}$$
Notice that the function $x\mapsto\sqrt[4]{\frac{x^4}{3}+4}$ is increasing.
So for $x>y$ the LHS is positive and the RHS is negative. Similarly, for $y<x$ the signs of the two expressions are opposite.
So we can only find a real solution for $x=y$, which gives us
$$2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}}x\tag{4}$$
From this we get
$$16\left(\frac{x^4}{3}+4\right)=\left(1+\sqrt{\frac{3}{2}}x\right)^4\tag{5}$$
This is a quartic equation. In theory, this can be done by hand, but it will be very probably quite messy. WolframAlpha returns this. (One of the solution, according to WA, is $\sqrt6$.)
