This is to continue the discussion in the comments. I know it should be put as a comment rather than an answer, but the system does not allow it due to the length. Please do not remove it.

Actually, this might work.

Let $K$ and $L$ be two fields of characteristic zero an such that there exists an embedding $K\hookrightarrow L$. Let $h_K : X_{K} \to {\mathbb{A}}^m_{K}$ be a log resolution over $K$ with numerical data ${(N_i,\nu_i)}_{i\in I}$, so

$$ div(h_{K}^*(\mathcal{I}(f)))=\sum_{i\in I} (\nu_i-1) E_i $$

and

$$ div(h_{K}^*(dx_1\wedge\dots\wedge dx_m))=\sum_{i\in I} N_i E_i. $$

Consider the commutative diagramme

$\require{AMScd}$

\begin{CD} X_{L} @> h_{L} >> {\mathbb{A}}^m_{L} \\ @V \pi_X V V @VV \pi_{\mathbb{A}} V\\ X_{K} @>> h_{K} > {\mathbb{A}}^m_{K}. \end{CD}

For every $i\in I$, we can possibly further decompose $\pi_X^{*} E_i$ into irreducible components. Write

$$ \pi_X^{*}E_i=\sum_{j\in J_i} a_jE_j', \qquad J:=\bigcup_{i\in I} J_i. $$

Then

$$ div(h_{L}^*(\mathcal{I}(f)))=\sum_{i\in I} \sum_{j\in J_i} (\nu_i-1)a_j E'_j $$

and

$$ div(h_{L}^*(dx_1\wedge\dots\wedge dx_m))=\sum_{i\in I} \sum_{j\in J_i} N_i a_j E'_j. $$

Now the inequality follows by noticing that

$$ \frac{(\nu_i-1)a_j+1}{N_ia_j}=\frac{\nu_i}{N_i}-\frac{1}{N_i}\Big(1-\frac{1}{a_j}\Big)\le \frac{\nu_i}{N_i}. $$

CORRECTION ON THE PREVIOUSLY GIVEN ANSWER

As remarked in the comments, Hironaka's construction behaves well under extension of the base field, that is a log-resolution for $(X_L,D_L)$ can be obtained via base change from a log-resolution of $(X_K,D_K)$ for any field extension $K\hookrightarrow L$. Also, an irreducible smooth divisor after base change is still smooth (although not necessarily irreducible). Since the irreducible components of a SNC divisor are smooth by definition, it follows that the numerical data of the resolution are left the same after base change (apart for possible repetitions). Therefore $lct(X_K,D_K)=lct(X_L,D_L)$.

Notation: $K$ is a field of characteristic zero, $(X_K,D_K)$ is a pair of a smooth variety over $K$ together with an effective non-zero divisor on it.

This is to continue the discussion in the comments. I know it should be put as a comment rather than an answer, but the system does not allow it due to the length. Please do not remove it.

Actually, this might work.

Let $K$ and $L$ be two fields of characteristic zero an such that there exists an embedding $K\hookrightarrow L$. Let $h_K : X_{K} \to {\mathbb{A}}^m_{K}$ be a log resolution over $K$ with numerical data ${(N_i,\nu_i)}_{i\in I}$, so

$$ div(h_{K}^*(\mathcal{I}(f)))=\sum_{i\in I} (\nu_i-1) E_i $$

and

$$ div(h_{K}^*(dx_1\wedge\dots\wedge dx_n))=\sum_{i\in I} N_i E_i. $$

Consider the commutative diagramme

$\require{AMScd}$

\begin{CD} X_{L} @> h_{L} >> {\mathbb{A}}^m_{L} \\ @V \pi_X V V @VV \pi_{\mathbb{A}} V\\ X_{K} @>> h_{K} > {\mathbb{A}}^m_{K}. \end{CD}

For every $i\in I$, we can possibly further decompose $\pi_X^{*} E_i$ into irreducible components. Write

$$ \pi_X^{*}E_i=\sum_{j\in J_i} a_jE_j', \qquad J:=\bigcup_{i\in I} J_i. $$

Then

$$ div(h_{L}^*(\mathcal{I}(f)))=\sum_{i\in I} \sum_{j\in J_i} (\nu_i-1)a_j E'_j $$

and

$$ div(h_{K}^*(dx_1\wedge\dots\wedge dx_n))=\sum_{i\in I} \sum_{j\in J_i} N_i a_j E'_j. $$

Now the inequality follows by noticing that

$$ \frac{(\nu_i-1)a_j+1}{N_ia_j}=\frac{\nu_i}{N_i}-\frac{1}{N_i}\Big(1-\frac{1}{a_j}\Big)\le \frac{\nu_i}{N_i}. $$

This is to continue the discussion in the comments. I know it should be put as a comment rather than an answer, but the system does not allow it due to the length. Please do not remove it.

Actually, this might work.

Let $K$ and $L$ be two fields of characteristic zero an such that there exists an embedding $K\hookrightarrow L$. Let $h_K : X_{K} \to {\mathbb{A}}^m_{K}$ be a log resolution over $K$ with numerical data ${(N_i,\nu_i)}_{i\in I}$, so

$$ div(h_{K}^*(\mathcal{I}(f)))=\sum_{i\in I} (\nu_i-1) E_i $$

and

$$ div(h_{K}^*(dx_1\wedge\dots\wedge dx_m))=\sum_{i\in I} N_i E_i. $$

Consider the commutative diagramme

$\require{AMScd}$

\begin{CD} X_{L} @> h_{L} >> {\mathbb{A}}^m_{L} \\ @V \pi_X V V @VV \pi_{\mathbb{A}} V\\ X_{K} @>> h_{K} > {\mathbb{A}}^m_{K}. \end{CD}

For every $i\in I$, we can possibly further decompose $\pi_X^{*} E_i$ into irreducible components. Write

$$ \pi_X^{*}E_i=\sum_{j\in J_i} a_jE_j', \qquad J:=\bigcup_{i\in I} J_i. $$

Then

$$ div(h_{L}^*(\mathcal{I}(f)))=\sum_{i\in I} \sum_{j\in J_i} (\nu_i-1)a_j E'_j $$

and

$$ div(h_{L}^*(dx_1\wedge\dots\wedge dx_m))=\sum_{i\in I} \sum_{j\in J_i} N_i a_j E'_j. $$

Now the inequality follows by noticing that

$$ \frac{(\nu_i-1)a_j+1}{N_ia_j}=\frac{\nu_i}{N_i}-\frac{1}{N_i}\Big(1-\frac{1}{a_j}\Big)\le \frac{\nu_i}{N_i}. $$