Question 806516
In any triangle, sum of any two sides is always greater than the third.


Consider the triangle ABN:
AN + BN > AB _______________ (1)


Consider the triangle BCN:
BN + CN > BC _______________ (2)


Consider the triangle ACN:
AN + CN > AC _______________ (3)


Adding both sides of (1), (2) and (3) we have

AN + BN + BN + CN + CN + AN > AB + BC + AC
2(AN + BN + CN) > AB + BC + AC
AN + BN + CN > (AB + BC + AC)/2


Hence proved