I am new to tropical geometry with addition is the $\max$ and multiplication the usually $+$ and I am trying to understand the translation of polynomials from the tropical setting to the classical setting. For example the abstract structure $$a\odot x^2\oplus b\odot x\oplus c= \max\lbrace 2x+a,x+b,c\rbrace$$ is clear but I am struggeling with the inverse way, if there is the multiplicative inverse involved for the reals. For example what is in tropical notation
$$p(x,y)=2\max\lbrace x,y\rbrace-(x+y)=\max\lbrace x,y\rbrace+\max\lbrace -x,-y\rbrace=?$$
I do not manage to resolve the negative signs here. Intuitively, it should be the analog of
$$q(x,y)=\frac{2(x+y)}{xy}$$ but this is not a polynomial.
Since the tropical multiplication is nothing else than the usual addition, it forms an abelian group. In consequence, each element possesses an inverse element with respect to multiplication, namely $x^{\odot-1} = -x$. N.B. : more generally, a tropical power can be defined as $x^{\odot n} = x \odot \ldots \odot x = n \cdot x$. You might introduce a new symbol, such as a circled division, in order to represent the tropical division.
Let's work on the first part of $p(x,y)$; one has : $$ p_1(x,y) := 2\max(x,y) = (x \oplus y)^{\odot2} = (x \oplus y) \odot (x \oplus y) = x^{\odot2} \oplus (x \odot y) \oplus (y \odot x) \oplus y^{\odot2} $$ where we used the distributivity of tropical multiplication over tropical addition. Now, the middle expression with the mixed products can be simplified thanks to the commutativity of tropical multiplication, i.e. $x \odot y = y \odot x$, as well as the idempotency of tropical addition, i.e. $a \oplus a = \max(a,a) = a$, so that $$ 2\max(x,y) = x^{\odot2} \oplus (x \odot y) \oplus y^{\odot2} $$
The second part of $p(x,y)$ is simply translated as $p_2(x,y) := -(x+y) = (x \odot y)^{\odot-1} = x^{\odot-1} \odot y^{\odot-1}$. It is then gathered together with the first part of $p(x,y)$ as follows : $$ p = p_1 + p_2 = p_1 \odot p_2 = \left(x^{\odot2} \oplus (x \odot y) \oplus y^{\odot2}\right) \odot x^{\odot-1} \odot y^{\odot-1} = (x \odot y^{\odot-1}) \oplus 0 \oplus (x^{\odot-1} \odot y), $$ where we used the distibutivity of $\odot$ over $\oplus$ again and the fact that $0$ is the identity element of $\odot$. The last expression can be considered as a standard bivariate tropical (Laurent) polynomial, which is indeed the tropical analog of $\frac{(x+y)^2}{xy} = xy^{-1} + 1 + x^{-1}y$.