In triangle ABC, given below, P is a point inside △ABC.

Prove that :

AP + BP + CP > $\dfrac{1}{2}$(AB + BC + CA)

Given: P is a point inside △ ABC.

To prove: AP + BP + CP > $\dfrac{1}{2}$(AB + BC + CA)

Proof: Using the Triangle Inequality Theorem, which states that the sum of the lengths of any two sides of a triangle is greater than the third side:

In Δ PAB:

AP + BP > AB ……………….(1)

Similarly, in Δ PBC:

BP + CP > BC ……………….(2)

Similarly, in Δ PAC:

AP + CP > AC ……………….(3)

Now, adding the equations (1), (2) and (3):

AP + BP + BP + CP + AP + CP > AB + BC + AC

⇒ 2AP + 2BP + 2CP > AB + BC + AC

⇒ 2(AP + BP + CP) > AB + BC + AC

⇒ AP + BP + CP > $\dfrac{1}{2}$(AB + BC + AC)

Hence, AP + BP + CP > $\dfrac{1}{2}$ (AB + BC + AC).