I'm given that $$\Lambda(\lambda) =\log M(\lambda) \leq \log( \frac{b-\bar{x}}{b-a} e^{\lambda a} + \frac{ \bar{x}-a}{b-a} e^{\lambda b}), \quad \bar{x}\in [a,b].$$
I'm trying to show that $$\Lambda^*(x) = \sup_{\lambda} \lambda x - \Lambda(\lambda) \geq H(\frac{x-a}{b-a} | \frac{\bar{x}-a}{b-a})$$ for $x\in [a,b]$, where $$H(p|p_0) = p \log (p/p_0) + (1-p) \log ((1-p)/(1-p_0)).$$
So far I've tried $$\Lambda^*(x) \geq \sup_{\lambda} \lambda x - \log( \frac{b-\bar{x}}{b-a} e^{\lambda a} + \frac{ \bar{x}-a}{b-a} e^{\lambda b}),$$
and maximizing the RHS by taking derivative of $\lambda x - \log( \frac{b-\bar{x}}{b-a} e^{\lambda a} + \frac{ \bar{x}-a}{b-a} e^{\lambda b})$ w.r.t $\lambda$ and setting to $0$. Then I got $\lambda^* = \frac{\log[\frac{(x-b)(a-z)}{(x-a)(b-z)}]}{a-b}$.
However, I get a value different from $H(\frac{x-a}{b-a} | \frac{\bar{x}-a}{b-a})$ when I plug in $\lambda^* = \frac{\log[\frac{(x-b)(a-z)}{(x-a)(b-z)}]}{a-b}$ into the RHS.
Any thoughts? I don't know what else I can do here.