Critical point exponents inequalities - The Rushbrooke inequality

 P: 316 The Rushbrooke inequality: $$H=0, T\rightarrow T_c^-$$ $$C_H \geq \frac{T\{(\frac{\partial M}{\partial T})_H\}^2}{\chi_T}$$ $$\epsilon=\frac{T-T_c}{T_c}$$ $$C_H \sim (-\epsilon)^{-\alpha'}$$ $$\chi_T \sim (-\epsilon)^{-\gamma'}$$ $$M \sim (-\epsilon)^{\beta}$$ $$(\frac{\partial M}{\partial T})_H \sim (-\epsilon)^{\beta-1}$$ $$(-\epsilon)^{-\alpha'} \geq \frac{(-\epsilon)^{2\beta-2}}{(-\epsilon)^{-\gamma'}}$$ and we get Rushbrooke inequality $$\alpha'+2\beta+\gamma' \geq 2$$ My only problem here is first step $$C_H \geq \frac{T\{(\frac{\partial M}{\partial T})_H\}^2}{\chi_T}$$ we get this from identity $$\chi_T(C_H-C_M)=T\alpha_H^2$$ But I don't know how?
 P: 316 Any idea? Problem is with $$\chi_T(C_H-C_M)=T\alpha^2_H$$ from that relation we obtain $$C_H=\frac{T\alpha^2_H}{\chi_T}+C_M$$ So if $$C_M>0$$ than $$C_H>\frac{T\alpha^2_H}{\chi_T}$$ Equality is in the game if and only if $$C_M=0$$. Or when $$(\frac{\partial^2 F}{\partial T^2})_M=0$$. Is that possible? F is Helmholtz free energy. Can you tell me something more about that physically?