Let $ABC$ be a triangle with centroid $G$. A perpendicular line from $G$ to the line $BG$ intersects the parallel through $A$ to the line $BC$ in $D$. Prove that $AC\cdot BD\geq 2\cdot area[AGBD]$.
So far, from the properties of the centroid and similarity of triangles I have noticed that $area[BGD]=4\cdot area[GHF]$, $area[AGD]=4\cdot area[GEH]$, $BD=2FH$ and $AC=2FC$ - see the figure. The points $E$ and $F$ are the midpoints of $BC$ and $AC$, and $H$ is the intersection of the lines $BC$ and $DG$. Thus, the inequality is equivalent with $CF\cdot FH\geq 2\cdot area[GEHF]$. I have still not used the perpendicularity of $DG$ on the median $BG$.
There is a solution below. I do not include all the details simply because we will be using very simple facts about medians, similar triangles, and the law of sine. Let's assume $\angle GBC=B_1$. Then, we have:
$$\frac{\frac{AC}{2}}{\sin B_1}=\frac{BF}{\sin C} \implies AC=\frac{2BF\times\sin B_1}{\sin C}, \\ BD= \sqrt {BG^2+DG^2}=\sqrt {BG^2+(2GH)^2}=\sqrt{(\frac{2}{3}BF)^2+(2 \times \frac{2}{3} \times BF \times \tan B_1)^2} \\ \implies \\ BD \times AC=\frac{4}{3}BF^2\sqrt{1+4\tan^2 B_1} \times \frac{\sin B_1}{\sin C}.$$
On the other hand,
$$2 \times (S_{\triangle AGD}+S_{\triangle DGB})=4 \times (2\times (S_{\triangle GFH}+S_{\triangle GEH})) \\ =4 \times (2\times (S_{\triangle GFH}+S_{\triangle GHB}-S_{\triangle GEB})) \\=4 \times ((\frac{BF}{3}\times \frac{2BF}{3}\times \tan B_1)+(\frac{2BF}{3}\times\frac{2BF}{3} \times \tan B_1-\frac{2BF}{3} \times \sin B_1 \times \frac{BC}{2})) \\=4 \times (\frac{2BF^2}{3}\times \tan B_1-\frac{BF}{3}\times \sin B_1 \times BC).$$
However, $\frac{BC}{BF}=\frac{\sin (B_1+C)}{\sin C}$. So,
$$2 \times (S_{\triangle AGD}+S_{\triangle DGB})=4 \times (\frac{2BF^2}{3}\times \tan B_1-\frac{BF^2 \sin (B_1+C) \sin B_1}{3 \sin C}).$$
Hence, we just need to show:
$$\sqrt{1+4\tan^2 B_1} \times \frac{\sin B_1}{\sin C} \geq 2\tan B_1-\frac{ \sin (B_1+C) \sin B_1}{ \sin C},$$
or equivalently,
$$\sqrt{1+4\tan^2 B_1} \times \frac{1}{\sin C} \geq \frac{2}{\cos B_1}-\frac{ \sin (B_1+C)}{ \sin C}=\frac{2\sin C-\cos B_1 \sin (B_1+C)}{\cos B_1 \sin C}. \ \ \ \ (*)$$
Now, note that:
$$|2\sin C-\cos B_1 \sin (B_1+C)|=|\sin C(2-\cos^2B_1)-\cos C (\sin B_1 \cos B_1)| \\ \leq (\sin ^2C +\cos ^2 C)^{\frac{1}{2}}((2-\cos^2B_1)^2+(\sin B_1 \cos B_1)^2)^{\frac{1}{2}}=\sqrt {\cos ^2 B_1+4\sin ^2 B_1} \ \ (**)\\ \implies |\frac{2\sin C-\cos B_1 \sin (B_1+C)}{\cos B_1 \sin C}| \leq \frac{\sqrt {\cos ^2 B_1+4\sin ^2 B_1}}{\cos B_1 \sin C}=\sqrt{1+4\tan^2 B_1} \times \frac{1}{\sin C}.$$
We are done.
Note $1$: We have assumed that $\cos B_1 >0 \ (0<B_1<90 ^{\circ}).$ Otherwise, the diagram will be entirely different, as well as the computations.
Note $2$: In $(**)$ the Cauchy–Schwarz inequality is used.