I'm reading the proof of Lemma 1 in Chapter 17 from Cassels' Lectures on Elliptic Curves. It says that
if we have homogeneous polynomials $D,E\in\mathbb Z[U_0,U_1]$ of degree $n$, and $u=(u_0:u_1)$ is a point on the rational projective line, and $D(u), E(u)$ do not both vanish, and furthermore, the resultant of $D$ and $E$ is not zero, then there exists a $\gamma>0$, independent of $u$, such that $$ H(D(u),E(u))\geq\gamma H(u)^n, $$ where $H$ is the height function.
The height function is defined as follows: for a projective point $(x:y)\in\mathbb P^1(\mathbb Q)$, such that $x,y\in\mathbb Z$ and $\gcd(x,y)=1$, we have $H(x,y)=\operatorname{max}(\vert x\vert,\vert y\vert)$.
By virtue of the resultant, there exist $L_j, M_j\in\mathbb Z[U_0,U_1]$ of degree at most $n-1$, such that $L_j D+M_j E=RU_j^{2n-1}$ ($j=0,1$). It follows then that $\gcd(D(u),E(u))\vert R$. Furthermore, for some $c>0$, we have $$ \vert L_j(u)\vert, \vert M_j(u)\vert \leq c(\max(\vert u_0\vert,\vert u_1\vert))^{n-1}, $$ for $j=0,1$.
Cassels says that it follows now that the the inequality in the conclusion holds for $\gamma=\vert R\vert/2c$. I don't see how they reached this conclusion. Could someone clarify this final step for me?
If you're fine with $\gamma=1/2c$, here's how to tie it together. The idea can be found in Silverman p. 223, where it is done for specific polynomials.
First off, we have $\delta:=\gcd(D(u),E(u))\vert R$, from which it follows that $\delta\leq |R|,$ and hence $$H(D(u),E(u))\geq \frac{\max\{|D(u)|,|E(u)|\}}{|R|}.$$ Furthermore, we know that $$ \max\{\vert L_j(u)\vert, \vert M_j(u)\vert\} \leq c(\max(\vert u_0\vert,\vert u_1\vert))^{n-1} $$ and from the resultant equation and the triangle inequality we see that $$ |Ru_j^{2n-1}|\leq 2\max\{|L_j(u)|,|M_j(u)|\}\max\{|D(u)|,|E(u)|\} $$ for $j=0,1$. Combining these two yields $$ \max\{|Ru_0^{2n-1}|,|Ru_1^{2n-1}|\}\leq 2c\max\{|u_0|,|u_1|\}^{n-1}\max\{|D(u)|,|E(u)|\}, $$ and we can cancel $\max\{|u_0|,|u_1|\}^{n-1}$ on both sides to get $$ |R|\max\{|u_0|,|u_1|\}^n\leq 2c\max\{|D(u)|,|E(u)|\}, $$ in other words $$ \max\{|D(u)|,|E(u)|\}\geq \frac{|R|}{2c}H(u)^n. $$ Putting it all together, we get $$ H(D(u),E(u))\geq \frac{\max\{|D(u)|,|E(u)|\}}{|R|}\geq \frac{1}{2c}H(u)^n. $$