Prove the inequality $u^{\alpha}v^{1-\alpha} \leq \alpha u + (1-\alpha)v $

Question

I'm trying to prove the inequality highlighted by the red line in the picture, but I do not know how. Please provide detail as much as you can, thank you!

For a minimal family, the sufficient statistic $\mathbf{T}$ is also minimal sufficient. For a proof, see Lehmann: Theory of Point Estimation, Example 5.9, pp. 43-44

If we parametrize the family using $\eta=c(\theta),$ this is called the natural parametrization (or the canonical parametrization). We then write $$ f(\mathbf{x}, \mathbf{\eta})=\exp \left(\mathbf{\eta}^{T} \mathbf{T}(\mathbf{x})-A(\mathbf{\eta})\right) h(\mathbf{x}) $$ where $$ A(\mathbf{\eta})=\log \int_{E} \exp \left(\mathbf{\eta}^{T} \mathbf{T}(\mathbf{x})\right) h(\mathbf{x}) d \mathbf{x} $$ The natural parameter space is $\mathbf{H}=\{\boldsymbol{\eta}: A(\boldsymbol{\eta})<\infty\} .$

Theorem 1 : H is a convex set.

Proof: Let $0<\alpha<1$ and take $\eta$ and $\eta_{1}$ in $\mathbf{H}$. Write $$ A\left(\alpha \eta+(1-\alpha) \eta_{1}\right)=\log \int_{E}\left(\exp \left(\eta^{T} \mathbf{T}(\mathbf{x})\right) h((\mathbf{x}))\right)^{\alpha}\left(\exp \left(\mathbf{\eta}_{1}^{T} \mathbf{T}(\mathbf{x})\right) h((\mathbf{x}))\right)^{1-\alpha} d \mathbf{x}$$ But $\underline{u^{\alpha} v^{1-\alpha} \leq \alpha u+(1-\alpha) v}$ (take logarithms of both sides and use the fact that the logarithm is a concave function), whence $$A\left(\alpha \eta+(1-\alpha) \eta_{1}\right) \leq \alpha A(\eta)+(1-\alpha) A\left(\eta_{1}\right)<\infty$$

Answer 1

For $u$ and $v$ are positives and $0\leq \alpha\leq1$ it's just AM-GM because $\alpha+1-\alpha=1.$

The AM-GM it's the following.

Let $\alpha_i\geq0$ such that $\sum\limits_{cyc}\alpha_i=1$ and $x_i>0$. Prove that: $$\alpha_1x_1+\alpha_2x_2+...+\alpha_nx_n\geq x_1^{\alpha_1}x_2^{\alpha_2}...x_n^{\alpha_n},$$

which is just Jensen for a $\ln$ function: $$\ln(\alpha_1x_1+\alpha_2x_2+...+\alpha_nx_n)\geq\alpha_1\ln{x_1}+\alpha_2\ln{x_2}+...+\alpha_n\ln{x_n}.$$

Answer 2

The inequality is a consequence of the fact that logarithm is a concave function on $(0,\infty)$.

The second derivative of $\log x$ is $-\frac 1{x^{2}} <0$. Hence $\log x$ is concave. Hence $\log (\alpha u+(1-\alpha) v) \geq \alpha \log u+(1-\alpha) \log v$. Now take exponential on both sides.