We use the following auxilliary statement.

**Lemma.** If \(A,B\) and \(C\) are pairwise
coprime polynomials with \(A+B=C\), then the degree of each of
them is less than the number of different zeroes of the polynomial
\(ABC\).

**Proof.** Let
\[A(x)=\prod_{i=1}^k(x-p_i)^{a_i},\quad B(x)=\prod_{i=1}^l(x-q_i)^{b_i}
,\quad C(x)=\prod_{i=1}^m(x-r_i)^{c_i}.\]
Writing the condition \(A+B=C\) as \(A(x)C(x)^{-1}+B(x)C(x)^{-1}=1\)
and differentiating it with respect to \(x\) gives us
\[A(x)C(x)^{-1}\left(\sum_{i=1}^k\frac{a_i}{x-p_i}-
\sum_{i=1}^m\frac{c_i}{x-r_i}\right)=
-B(x)C(x)^{-1}\left(\sum_{i=1}^l\frac{b_i}{x-q_i}-
\sum_{i=1}^m\frac{c_i}{x-r_i}\right),\] from which we see that
\(A(x)/B(x)\) can be written as a quotient of two polynomials od
degrees not exceeding \(k+l+m-1\). Our statement now follows from
the fact that \(A\) and \(B\) are coprime.

Apply this statement on polynomials \(P^a,Q^b,R^c\). Each of their
degrees \(a\deg P\), \(b\deg Q\), \(c\deg R\) is less than \(\deg P+\deg
Q+\deg R\) and hence \(\frac1a > \frac{\deg P}{\deg P+\deg Q+\deg R}\),
etc. Summing up yields the desired inequality.

** Corollary.** ``The Last Fermat's theorem_quot_ for polynomials.