Let us give some explicit formulae for the lower-bound above, in some particular cases. We consider $X=\mathbf{P}^{1}$ over ${\mathbf{Q}}$ and the metrized line bundle $\overline{\mathscr{O}(1)}_{\mathrm{W}}$. Let $\varphi$ and $\psi$ be polynomials with integral coefficients, of degrees $\ell$ and $m$ respectively ; let us pose $\overline{L}=\varphi^{*}\overline{\mathscr{O}(1)}_{\mathrm{W}}$, $\overline{M}=\psi^{*}\overline{\mathscr{O}(1)}_{\mathrm{W}}$. The line bundle $\overline{L}^{m}\otimes\overline{L}^{-\ell}$ is trivial and its metric is given by a family of functions $(f_{v})$. Since $\varphi$ and $\psi$ have integral coefficients, $f_{v}=0$ at all finite places. Moreover, since $g_{\overline{L}}(x)=\log\max(\left|{\varphi(x)}\right|,1)$ and $g_{\overline{M}}(x)=\log\max(\left|{\psi(x)}\right|,1)$ are the Green functions for the divisors $\ell[\infty]$ and $m[\infty]$ respectively, one has

$$f_{\infty}(x)=\log\frac{\max(\left|{\varphi(x)}\right|^{m},1)}{\max(\left|{\psi(x)}\right|^{\ell},1)}.$$

Then,

$$\mathop{\mathrm{d}\mathrm{d}^{c}}f_{\infty}=\frac{m}{2\pi}\mathrm{d}\operatorname{Arg}\varphi(x)\wedge\delta_{\left|{\varphi(x)}\right|=1}-\frac{\ell}{2\pi}\mathrm{d}\operatorname{Arg}\psi(x)\wedge\delta_{\left|{\psi(x)}\right|=1}$$

From this, we deduce that

	$\displaystyle\mathscr{D}(f_{\infty})$	$\displaystyle=\frac{\ell m}{2\pi}\left(\int_{\left|{\psi(x)}\right|=1}\log\max(\left|{\varphi(x)}\right|,1)\mathrm{d}\operatorname{Arg}\psi(x)\right.$
		$\displaystyle\qquad{}\left.+\int_{\left|{\varphi(x)}\right|=1}\log\max(\left|{\psi(x)}\right|,1)\mathrm{d}\operatorname{Arg}\varphi(x)\right),$

the two others terms vanishing. In fact, Stokes’s formula implies that the two terms within the parentheses in the previous formula are equal and we have

$$\mathscr{D}(f_{\infty})=\frac{\ell m}{\pi}\int_{\left|{\varphi(x)}\right|=1}\log\max(\left|{\psi(x),1}\right|)\mathrm{d}\operatorname{Arg}\varphi(x).$$

The simplest case to study is for $\varphi(x)=x^{\ell}$. Then,

$$\mathscr{D}(f_{\infty})=\frac{\ell m}{\pi}\int_{0}^{2\pi}\log\max(\left|{\psi(e^{i\theta})}\right|,1)\,\mathrm{d}\theta$$

is $2\ell m$ times the logarithm of the variant $\mathrm{M}^{+}(\psi)$ of the Mahler measure of $\psi$ :

$$\mathrm{M}^{+}(\psi)=\exp\left(\frac{1}{2\pi}\int_{0}^{2\pi}\log\max(\left|{\psi(e^{i\theta})}\right|,1)\,\mathrm{d}\theta\right).$$

In fact, Jensen’s formula implies that

$$\mathrm{M}^{+}(\psi)=\exp\left(\frac{1}{(2\pi)^{2}}\int_{0}^{2\pi}\log\left|{\psi(e^{i\theta_{1}})-e^{i\theta_{2}}}\right|\,\mathrm{d}\theta_{1}\mathrm{d}\theta_{2}\right)$$

is the Mahler measure $\mathrm{M}(\psi(x)-y)$ of the 2-variables polynomial $\psi(x)-y$.

Consequently, except for finitely many exceptions, any algebraic point $x\in{\mathbf{P}}^{1}(\overline{{\mathbf{Q}}})$ satisfies

$$\ell h(x)+h(\psi(x))\geq\frac{1}{\ell+m}\log\mathrm{M}(\psi(x)-y).$$

For $\ell=1$ and $\psi(x)=1-x$, we obtain that up to finitely many exceptions,

$$h(x)+h(1-x)\geq\frac{1}{2}\log\mathrm{M}(1-x-y)\approx 0.161538,$$

In that particular case, Zagier [56] has proved a much more precise result : except for 5 explicit points in ${\mathbf{P}}^{1}$,

$$h(x)+h(1-x)\geq\frac{1}{2}\log(\frac{1+\sqrt{5}}{2})\approx 0.240606.$$

Observe also that if $(x_{j})$ is a sequence of points such that $h(x_{j})\rightarrow 0$, Theorem 3.3.1 implies that $h(1-x)\rightarrow\log\mathrm{M}(1-x-y)\approx 0.323076$.