Assume that the first rectangle $EFGH$ can be placed within the second $ABCD$. Translate it and rotate it within the second rectangle until their centers coincide and at least one pair of opposite vertices of the first rectangle lie on the sides of the second. Now, fix this pair of vertices and rotate the other diagonal of the first rectangle until all four vertices lie on the sides of the second. Because $b > d$, $b> c$, $x>c$ and $x>d$ so the area of the inner rectangle has only increased (the rotation can't go too far). Call this rectangle $E'F'G'H'$, with $E'$ on $DA$, $F'$ on $AB$, etc. Note that because $b>c$ and $b>d$, we can assume that $E', F'$ are closer to $A$ than $D$, $B$, respectively, and similarly for $G'$, $H'$. Now, it is easy to compute that, if $x$ denotes the diagonal of the $EFGH$, of the "leftover" triangles of $ABCD$, two have sides $\frac{c+\sqrt{x^2-d^2}}{2}$, $\frac{d + \sqrt{x^2-c^2}}{2}$ and the other two have sides $\frac{c-\sqrt{x^2-d^2}}{2}$ and $\frac{d-\sqrt{x^2-d^2}}{2}$, so that the total area of all four triangles is: $\frac{cd + \sqrt{(x^2-c^2)(x^2-d^2)}}{2}$ Therefore the inner rectangle has area $ab \leq \frac{cd - \sqrt{(x^2-c^2)(x^2-d^2)}}{2}$ which is equivalent upon simplifying and squaring to the desired inequality. This procedure is reversible, so we're done.