Answer:
In\[\Delta AOB\], OA + OB > AB ....(i) Similarly, in ABOC, OB + OC > BC ....(ii) and in\[\Delta AOC\], OA + OC > AC ....(iii) By adding (i), (ii) and (iii), we get OA + OB + OB + OC + OA + OC > AB + BC +CA 20A + 20B + 20C > AB + BC + CA 2(OA+OB+OC)>AB+BC+CA
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