AD is the median of a ΔABC, prove that AB + BC + CA > 2AD (HOTS)

In ΔABD,

AB + BD > AD …(i)

(Sum of two sides of a triangle is greater than the third side)

Similarly, In ΔADC, we have

AC + DC > AD …(ii)

Adding (i) and (ii), we have

AB + BD + AC + DC > 2AD

⇒ AB + (BD + DC) + AC > 2AD

⇒ AB + BC + AC > 2AD

Hence, proved.