Question

P is a point in the exterior of ∆XYZ, show that PX+PY+PZ>1/2(XY+YZ+ZX)

Answer 1

Answer:

(PX + PY + PZ) > (1/2) (XY + YZ + ZX)

Step-by-step explanation:

in Δ PXY

PX + PY > XY  ( sum of two sides of triangle > third side)

in Δ PXZ

PX + PZ > XZ

in Δ PYZ

PY + PZ > YZ

Adding all three

PX + PY + PX + PZ + PY + PZ > XY + XZ + YZ

=> 2(PX + PY + PZ) > XY + YZ + ZX

=> (PX + PY + PZ) > (1/2) (XY + YZ + ZX)

QED

Proved