If $O$ is a point inside $\triangle ABC$,Prove: $$\frac{\overline{AB}+\overline{BC}+\overline{CA}}{2}<\overline{AO}+\overline{BO}+\overline{CO}<\overline{AB}+\overline{BC}+\overline{CA}$$
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Things I have done: I was able to proof second part $$Proof \space of \space second \space part\\\\ \begin{array}{ l| l } \hline Statement & Reasoning \\ \hline \overline{AB}+\overline{AE} > \overline{BE} & Triangle\space inequality \\ \overline{OE}+\overline{CE} > \overline{OC} & Triangle\space inequality \\ \overline{AB}+\overline{AC} > \overline{OC} + \overline{OB} & By \space statement\space 1 \space and \space 2 \\ \overline{AC}+\overline{BC} > \overline{OA} + \overline{OB} & Similar \space to \space statement\space 1 \space and \space 2 \space and \space 3 \\ \overline{AB}+\overline{BC} > \overline{OA} + \overline{OB} & Similar \space to \space statement\space 1 \space and \space 2 \space and \space 3 \\ \overline{AO}+\overline{BO}+\overline{CO}<\overline{AB}+\overline{BC}+\overline{CA} & By \space statement\space 3 \space and \space 4 \space and \space 5\\ \end{array}$$
For the first inequality which is between semiperimeter and sum of distances from vertices, I don't know what to do. I think my approach for the second inequality can't be applied to proving this one.
By the triangle inequality, $$ OA+OB>AB;\\ OA+OC>AC;\\ OB+OC>BC. $$ Summing these and dividing by 2 give you the first inequality.