Problem
Given two triangles $ABC$ and $A'B'C'$ where $a,b,c$ and $a',b',c'$ are the corresponding sides and $F, F'$ denotes the areas of the two triangles. Prove: \begin{equation} a^2a'^2+b^2b'^2+c^2c'^2\geqslant 16FF'. \end{equation}
Solution
I know that $ a^2\left(-a'^2+b'^2+c'^2\right)+b^2\left(a'^2-b'^2+c'^2\right)+c^2\left(a'^2+b'^2-c'^2\right)\geqslant 16FF',$ which is the Pedoe's inequality, and $\cot A \left(\cot B'+\cot C'\right)+\cot B \left(\cot A'+\cot C'\right)+\cot C \left(\cot A'+\cot B'\right)\geqslant 2$, which is an equivalent version of the Pedoe's inequality. Therefore, I was trying to solve this problem by transforming side lengths into cotangents. Because $$\cot A=\frac{\cos A}{\sin A}=\frac{b^2+c^2-a^2}{2bc}\cdot \frac{bc}{2F}=\frac{b^2+c^2-a^2}{4F}\,, \quad \text{(by the law of sines and the law of cosines)}$$ the inequality we desire is equivalent to $$\left(\cot B+\cot C\right)\left(\cot B'+\cot C'\right)+\left(\cot C+\cot A\right)\left(\cot C'+\cot A'\right)+\left(\cot A+\cot B\right)\left(\cot A'+\cot B'\right)\geqslant 4.$$ Expanding it and arrange terms gives us $$2\left(\cot A\cot A'+\cot B\cot B' +\cot C\cot C'\right) + M\geqslant 4$$ where $M=\cot A \left(\cot B'+\cot C'\right)+\cot B \left(\cot A'+\cot C'\right)+\cot C \left(\cot A'+\cot B'\right)$. We know $M\geqslant 2$, so I thought if I can prove $\cot A\cot A'+\cot B\cot B' +\cot C\cot C \geqslant 1$, the problem would be solved. However, this might not be true. After this, I got stuck.
Another inequality related to this problem is $$a^2x+b^2y+c^2z\geqslant 4F\sqrt{xy+yz+zx}\quad \forall x,y,z\in \mathbb{R}^+.\quad\text{(Oppenheim's inequality)} $$ If we set $x=a'^2, y=b'^2, z=c'^2$, we can get $$a^2a'^2+b^2b'^2+c^2c'^2\geqslant 4F\sqrt{\frac{(2F')^2}{\sin A'}+\frac{(2F')^2}{\sin B'}+\frac{(2F')^2}{\sin C'}}=8FF'\sqrt{\csc^2 A'+\csc^2 B' +\csc^2 C'}.$$ And by Jensen's inequality, we can get $$a^2a'^2+b^2b'^2+c^2c'^2\geqslant 8FF' \cdot \sqrt{3\csc^2\left(\frac{A'+B'+C'}{3}\right)}=16FF'. \square $$ Notice that when $a'=b'=c'$ our inequality is essentially Weitzenböck's inequality, so when both triangles our equilateral triangles, we get an equality. I guess this inequality can also be directly derived from Pedoe's inequality (maybe without trig), but I don't know how to prove this only by Pedoe's. I'm also willing to see other perspectives (maybe pure geometry?...) to this problem.
Note/My Opinion: Since Pedoe's inequality had been incorporated in the proof of Oppenheim's inequality, and the proof uses the trig form of it, I don't count my proof as an direct implication of the Pedoe's.
