Surface area of tetrahedron bigger than product of diagonals

Question

Let $ABCD$ be a tetrahedron with $AB=6,BC=9,CD=7$, and $DA=2$. Let $S$ be the surface area of the tetrahedron. Prove that $S>AC\cdot BD$.

By triangle inequality we have $AC

Answer 1

Let $AC=x$ and $BD=y$ and denote the semi-perimeters of the triangles $ABD,BCD,ACD,ABC$ by $p_1,p_2,p_3,p_4$ respectively. Then $$\begin{align} p_1&=4+\frac y2,\quad p_2=8+\frac y2\\ p_3&=\frac{9+x}2,\quad p_4=\frac{15+x}2 \end{align}$$ and by Heron's formula, the surface area of the tetrahedron is calculated from $$4S=\sqrt{-2025 + 106 x^2 - x^4} + \sqrt{-2025 + 234 x^2 - x^4}\\+ \sqrt{-1024 + 80 y^2 - y^4} + \sqrt{-1024 + 260 y^2 - y^4}$$ where you need to show that $S>xy$.

To obtain a lower-bound for $S$, note that $$S=f_1(x)+f_2(y)$$ which means $$\min S=\min_x f_1+\min_y f_2$$ So I simply ran the following code in Mathematica:

{Minimize[{f1[x], x > 0}, x], Minimize[{f2[y], y > 0}, y]}

the result shows that $S$ is minimized where $x=5,y=4$ and $$\min S=10\sqrt 2+6\sqrt 5>20$$ Also note that $xy$ is maximized when both $x$ and $y$ are maximum. Now the maximum value of $x$ in the domain of $f_1(x)$ is $9$ and maximum of $y$ in the domain of $f_2(y)$ is $8$. But $$f_1(9)+f_2(8)=18\sqrt 2+12\sqrt 5\approx 52.3<9\times 8$$ it looks like the statement is false