Does $\max.\mathbb{E}\left\{ \log (1+\frac{\alpha}{\beta} )\right\}$ and $\max.\mathbb{E}\left\{\frac{\alpha}{\beta}\right\}$ share the same solution?

I want to prove or disprove that the optimal solutions of $$\max_{\mathbf{x}}f(\mathbf{x})=\max_{\mathbf{x}}\mathbb{E}_{\mathbf{d}}\left\{ \log (1+\frac{\alpha}{\beta} )\right\}$$ and $$\max_{\mathbf{x}}g(\mathbf{x})=\max_{\mathbf{x}}\mathbb{E}_{\mathbf{d}}\left\{\frac{\alpha}{\beta}\right\}$$ are the same, where $$\alpha = a\left|\mathbf{x}^H\mathbf{d}\right|^2$$, $$\beta = \mathbf{x}^H\left(\mathbf{R}+b\mathbf{d}\mathbf{d}^H\right)\mathbf{x}$$.

$$\mathbf{R}\in\mathbb{C}^{n\times n},a>0,b>0$$ are known positive definite matrix and scalars, respectively. $$\mathbf{d}\in\mathbb{C}^{n\times 1}$$ is the Gaussian random vector with mean $$\mathbf{\mu}$$, and covariance matrix $$\mathbf{\Sigma}$$. $$\mathbb{E}_{\mathbf{d}}\left\{\cdot\right\}$$ denotes the expectation operation with recpect to $$\mathbf{d}$$. $$\left\{\cdot\right\}^H$$ denotes the conjugate transpose.

My thinking is that we first take the partial derivative of $$f(\mathbf{x})$$ and $$g(\mathbf{x})$$. That is $$\partial f(\mathbf{x}) = \mathbb{E}_{\mathbf{d}}\left\{\frac{\beta\partial\alpha-\alpha\partial\beta}{\alpha\beta+\beta^2}\right\} = \mathbb{E}_{\mathbf{d}}\left\{\frac{\frac{1}{\alpha}\partial\alpha-\frac{1}{\beta}\partial\beta}{1+\frac{\beta}{\alpha}}\right\},$$ and $$\partial g(\mathbf{x}) = \mathbb{E}_{\mathbf{d}}\left\{\frac{\beta\partial\alpha-\alpha\partial\beta}{\beta^2}\right\} = \mathbb{E}_{\mathbf{d}}\left\{\frac{\frac{1}{\alpha}\partial\alpha-\frac{1}{\beta}\partial\beta}{\frac{\beta}{\alpha}}\right\}.$$

From the optimal condition that $$\frac{\partial f(\mathbf{x})}{\partial\mathbf{x}} = \mathbf{0}$$ and $$\frac{\partial g(\mathbf{x})}{\partial\mathbf{x}} = \mathbf{0}$$, we finally have that

$$\mathbb{E}_{\mathbf{d}}\left\{\frac{\frac{a}{\alpha}\mathbf{x}^H\mathbf{d}\mathbf{d}^H- \frac{1}{\beta}\mathbf{x}^H\left(\mathbf{R}+b\mathbf{d}\mathbf{d}^H\right) }{1+\frac{\beta}{\alpha}} \right\} = \mathbf{0},$$ and $$\mathbb{E}_{\mathbf{d}}\left\{\frac{\frac{a}{\alpha}\mathbf{x}^H\mathbf{d}\mathbf{d}^H- \frac{1}{\beta}\mathbf{x}^H\left(\mathbf{R}+b\mathbf{d}\mathbf{d}^H\right) }{\frac{\beta}{\alpha}} \right\} = \mathbf{0}.$$

It can be observed that the above two equations have the same numerator. If there is no expectation operator, they will have the same solution. However, the trouble now is the denominator and the expectation operator. Can someone help with this? Thanks.