= Solution
The <scaling hypothesis for critical phenomena> says that the singular equilibrium <free-energy density> is a generalized homogeneous function of the thermal and field variables, after analytic backgrounds are removed. To derive the ordinary mean-field form, put $r=r_t t$, $u>0$ and neglect the higher powers near the transition. Rescale $m=\sqrt{r_t/u}\,|t|^{1/2}\psi$. Minimizing the quartic potential becomes
$$
A_s=\frac{r_t^2}{u}|t|^2\min_\psi\left[\frac{\operatorname{sgn}(t)}2\psi^2+\frac14\psi^4-\frac{\sqrt u}{r_t^{3/2}}\frac h{|t|^{3/2}}\psi\right].
$$
This is the <mean-field scalar free-energy scaling> form with $a=r_t^2/u$, $b=\sqrt u/r_t^{3/2}$. The two functions correspond to $t>0$ and $t<0$; the latter has two pure-phase branches and a cusp at zero field. The question's $>$ and $<$ subscripts distinguish these temperatures above and below $T_c$.
Differentiating with respect to $h$ gives $m=-\partial_hA_s\propto|t|^{1/2}$ on an ordered pure branch, and differentiating again gives $\chi=-\partial_h^2A_s\propto|t|^{-1}$. Thus the displayed scaling form independently yields $\beta=1/2$ and $\gamma=1$.
To allow anomalous powers, replace it by
$$
A_s=a|t|^{2-\alpha}f_\pm\left(\frac{bh}{|t|^\Delta}\right).
$$
Then $\beta=2-\alpha-\Delta$ and $\gamma=2\Delta-(2-\alpha)$. Taking $t\to0$ at fixed small $h$ gives $A_s\propto|h|^{(2-\alpha)/\Delta}$, so $1/\delta=(2-\alpha-\Delta)/\Delta=\beta/\Delta$. Eliminating $\Delta$ proves the <Rushbrooke scaling relation> and <Widom scaling relation>:
$$
\boxed{\alpha+2\beta+\gamma=2,\qquad\beta\delta=\Delta=\beta+\gamma.}
$$
These power-law relations assume the simple scaling form; marginal corrections or <dangerously irrelevant couplings> can require additional scaling variables.
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