I try to answer the following question
Let $F, G, H$ are three projective plane curves and $$\sum_P I(P,F\cap H)>\sum_Q I(Q,G\cap H) $$ Show that there exists an irreducible curve $B$ such that $$\sum_P I(P,F\cap H)-\sum_Q I(Q,G\cap H)=\sum_R I(R,B\cap H)$$
My attempt: It is obvious that these three curves do not share the common components. Therefore, according to the Bezout's theorem, we have $$\sum_P I(P,F\cap H)-\sum_Q I(Q,G\cap H)=\deg (H)(\deg(F)-\deg(G)).$$ So what we need to do is to find an irreducible curve in $\bf{P}^2$ which does not share the common components with $H$ and is of degree $\deg(F)-\deg(G)$.
Regardless of the "irreducible" condition, it is obvious that we can take the union of $(\deg(F)-\deg(G))$ projective lines. (And the existence of these projective lines is obvious geometrical intuitionally, but how to write it down?)
Furthermore, the union of several projective lines is definitely not irreducible. So how can we find an irreducible curve in $\bf{P}^2$ of a given degree?
Or we can think about the prime ideals in the polynomial ring $k[x,y,z]$? But I think there is no such a nice property to tell us the irreducible thing.
The set of non-zero homogeneous polynomials of degree $d$ in 3 variables form a projective space of dimension $N_d={d+2\choose 2}-1$. The set of polynomials which are product of two polynomials of degrees $e,f$ with $e+f=d$ is of dimension at most $N_e+N_f$, since it is the image of $\mathbb{P}^{N_e}\times\mathbb{P}^{N_f}\to\mathbb{P}^{N_d}$, under the multiplication map. An easy calculation shows $N_e+N_f<N_d$ and since there are only finitely many choices for $e, f$, both positive and $e+f=d$, we see that there is a non-empty open set in $\mathbb{P}^{N_d}$ consisting of irreducible polynomials of degree $d$.