Let $\mathcal{A}$ and $\mathcal{B}$ be two open bounded non-empty subsets of $\mathbb{R}^{2}$ and $\lambda \in ]0,1[$. Verify that $\lambda\mathcal{A} + (1-\lambda)\mathcal{B}$ is an open bounded subset of $\mathbb{R}^{2}$. Then show that $$V(\lambda\mathcal{A} + (1-\lambda)\mathcal{B}) \geq V(\mathcal{A})^{\lambda} V(\mathcal{B})^{1-\lambda}$$ To prove this inequality, you will use the following admitted result. For all $f \in C(\mathcal{A})$ and $g \in C(\mathcal{B})$, the function $h$ determined by: $$\forall Z \in \mathbb{R}^{2},\quad h(Z) = \sup\left\{f(X)^{\lambda} g(Y)^{1-\lambda} \,/\, X, Y \in \mathbb{R}^{2},\, Z = \lambda X + (1-\lambda) Y\right\}$$ defines a continuous function on $\mathbb{R}^{2}$.
Let $\mathcal{A}$ and $\mathcal{B}$ be two open bounded non-empty subsets of $\mathbb{R}^{2}$ and $\lambda \in ]0,1[$. Verify that $\lambda\mathcal{A} + (1-\lambda)\mathcal{B}$ is an open bounded subset of $\mathbb{R}^{2}$. Then show that
$$V(\lambda\mathcal{A} + (1-\lambda)\mathcal{B}) \geq V(\mathcal{A})^{\lambda} V(\mathcal{B})^{1-\lambda}$$
To prove this inequality, you will use the following admitted result. For all $f \in C(\mathcal{A})$ and $g \in C(\mathcal{B})$, the function $h$ determined by:
$$\forall Z \in \mathbb{R}^{2},\quad h(Z) = \sup\left\{f(X)^{\lambda} g(Y)^{1-\lambda} \,/\, X, Y \in \mathbb{R}^{2},\, Z = \lambda X + (1-\lambda) Y\right\}$$
defines a continuous function on $\mathbb{R}^{2}$.