A problem that everyone must try.. Let $H$ be the orthocentre of $ABC$. Then $AH=2R\cos A$ etc and $BC=2R\sin A$ etc (well-known). Note that $AH^2+BC^2-CH^2=4R^2(\cos^2A+\sin^2A-\cos^2C)=4R^2(1-\cos^2C)=$ $4R^2\sin^2C=AB^2$ The following equations hold due to Pythagoras' theorem: $AH^2+HD^2=AD^2$ (1) $BH^2+HD^2=BD^2$ (2) $CH^2+HD^2=CD^2$ (3) $BD^2+CD^2=BC^2$ (4) Now (2)+(3) imply $2HD^2=BC^2-CH^2-BH^2$ using (4). Therefore (1)+(2) imply $AD^2+BD^2=AH^2+BC^2-CH^2=AB^2$. So: $BD^2+CD^2=BC^2$ $AD^2+BD^2=AB^2$ $CD^2+AD^2=CA^2$ Adding these up and multiplying by 3 gives $6(AD^2+BD^2+CD^2)=3(AB^2+BC^2+CD^2)$. And clearly $3(AB^2+BC^2+CD^2)\ge (AB+BC+CD)^2$. (we have equality when $ABC$ is equilateral)