Let $a, b, \lambda \in (0, \infty)$ each be a D5407: Positive real number such that

(i)	\begin{equation} a + b + \sqrt{a^2 + b^2} = \lambda \end{equation}

Then

(1)	\begin{equation} \frac{1}{2} a b \leq \left( \frac{\lambda - \sqrt{a^2 + b^2}}{\sqrt{8}} \right)^2 \end{equation}
(2)	\begin{equation} \frac{1}{2} a b = \left( \frac{\lambda - \sqrt{a^2 + b^2}}{\sqrt{8}} \right)^2 \quad \iff \quad a = b \end{equation}