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  \[
    I_{\mathcal{P}} \left( \frac{p_1}{q_1},
    \frac{p_2}{q_2} \right) \ge \frac{\eps^2}{20}
    h(n_1 \log q_1 + n_2 \log q_2)
  .\]
  If $F$ vanishes to order $m$ at
  $\frac{p_1}{q_1}$, then $q_1^m$ divides the
  leading coefficient of $F$. In particular,
  $q_1^m \le H(F)$.

  Then
  \[
    I_F \left( \frac{p_1}{q_1}; \log q_1 \right)
    \le \log H(\mathcal{F}) \le \frac{10 hn_1 \log
    q_1}{C}
  \]
  \[
    I_G \left( \frac{p_2}{q_2}; \log q_2 \right)
    \le \log H(\mathcal{G}) \le \frac{10 hn_2 \log
    q_2}{C}
  \]
  If $C$ is sufficiently large in terms of $\eps$,
  then
  \[
    I_{\mathcal{P}} \left( \frac{p_1}{q_1},
    \frac{p_2}{q_2} \right) < I_{\mathcal{F}}
    \left( \frac{p_1}{q_1} \right) +
    I_{\mathcal{G}} \left( \frac{p_2}{q_2} \right)
  .\]
  A contradiction.
\end{proof}

Now we prove the lemma from earlier:

\begin{lemma*}
  Let $F^{(1)}, F^{(2)}, \ldots, F^{(h)}$ be
  $\Qbb$-linearly independent polynomials in
  $\Zbb[X]$. Then
  \[
    \begin{vmatrix}
      F^{(1)} & F^{(2)} & \cdots F^{(h)} \\
      F_1^{(1)} & F_1^{(2)} & \cdots & F_1^{(h)}
      \\
      \vdots & \vdots & \ddots & \vdots \\
      F_{h - 1}^{(1)} & F_{h - 1}^{(2)} & \cdots &
      F_{h - 1}^{(h)}
    \end{vmatrix}
    \neq 0
  .\]
  (Wronskian)
\end{lemma*}

\begin{proof}
  The statement does not change if we replace
  $F^{(j)}$ by $a F^{(i)} + b F^{(j)}$ for some
  $a, b \in \Qbb$ and $i \in \{1, \ldots, h\}$
  provided $b \neq 0$.

  Then we may assume: $F^{(i)} = X^{mi} +
  \text{lower order terms}$ and the $m_i$ are
  distinct.

  We will prove that:
  \[
    \begin{vmatrix}
      X^{m_1} & \cdots & X^{m_h} \\
      {m_1 \choose 1} X^{m_1 - 1} & \cdots & {m_h
      \choose 1} X^{mh - 1} \\
      \vdots & \ddots & \vdots \\
      {m \choose h - 1} X^{m_1 - h + 1} & \cdots
      & {mh \choose h - 1} X^{m_h - h + 1}
    \end{vmatrix}
    \neq 0
  .\]
  Then this is the leading term of th Wronskian,
  so this will prove the claim. The determinant is
  equal to:
  \[
    \begin{vmatrix}
      {m_1 \choose 0} & \cdots & {m_h \choose 0}
      \\
      \vdots & \ddots & \vdots \\
      {m_1 \choose h - 1} & \cdots & {m_h \choose
      h - 1}
    \end{vmatrix}
    \cdot X^M
  \]
  Supose to the contrary that a non-trivial linear
  combination of the rows is $(0, 0, \ldots, 0)$.
  Now the $i$-th row is a polynomial of degree $i
  - 1$evaluated at $m_1, \ldots, m_h$. Then the
  linear combination of the rows is a non-zero
  polynomial of degree $\le h - 1$ evaluated at
  $m_1, \ldots, m_h$.
\end{proof}

\begin{theorem*}
  Let $\alpha$ be an irrational, real algebraic
  number of degree $d \ge 2$. Then for all $\eps >
  0$, there exists $C = C(\alpha, \eps)$ such that
  \[
    \left| \alpha - \frac{p}{q} \right| > C
    q^{-\sqrt{2d} - \eps}
  ,\]
  for all $\frac{p}{q} \in \Qbb$.
\end{theorem*}

\begin{proof}
  Suppose to the contrary that there are
  infinitely many $\frac{p}{q}$ with
  \[
    \left| \alpha - \frac{p}{q} \right| <
    q^{-\sqrt{2d} - \eps}
  .\]
  Then fix $\eps_0 > 0$ sufficiently small in
  terms of $\alpha, \eps$ and let $C$ be the
  constant when the proposition is applied with
  $\eps_0$ in place of $\eps$.

  Now let $\frac{p_1}{q_1}, \frac{p_2}{q_2}$ be
  such that
  \[
    \left| \alpha - \frac{p_1}{q_1} \right|,
    \left| \alpha - \frac{p_2}{q_2} \right| <
    q^{-\sqrt{2d} - \eps}
  \]
  and
  \[
    \log q_1 > C \cdot \eps_0^{-1}
    \qquad \log q_2 > C \log q_1
  .\]
  We use \nameref{siegel} to construct $P(X_1,
  X_2)$ that vanishes at $(\alpha, \alpha)$ to
  high order.

  We choose $n_1, n_2 \in \Zbb$ such that
  \[
    n_1 \log q_1 \le n_2 \log q_2 \le n_1 \log q_1
    + \log q_1
  .\]
  We want a polynomial $P$ such that
  \[
    I_P(\alpha, \alpha) \ge \frac{n_1 \log q_1 +
    n_2 \log q_2}{\sqrt{2d} + \frac{\eps}{10}}
  .\]
  For this we need to estimate
  \begin{align*}
    |\{(i_1, i_2) \in \Zbb_{\ge 0}^2 : i_1 \log
    q_1 + i_2 \log q_2 \le I\}|
    &\le \frac{(I +
    \log q_1 + \log q_2)^2}{2 \log q_1 \log q_2}
    \\
    &\le \frac{(n_1 + 1)(n_2 + 1)}{(1 + \delta) d}
  \end{align*}
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/2c79cd0366a043ca.png}
  \end{center}
  This is because
  \[
    I \sim \frac{2n_1 \log q_1}{\sqrt{2d}} \sim
    \frac{2n_2 \log q_2}{\sqrt{2d}}
  \]
  so
  \[
    \frac{I^2}{2 \log q_1 \log q_2} \sim
    \frac{2n_1 \cdot 2n_2}{2 \cdot 2d}
    = \frac{n_1 n_2}{d}
  .\]
