Question

If $O$ is a point within triangle $ABC$, show that,

1. $AB+AC>OB+OC$
2. $AB+BC+CA>OA+OB+OC$
3. $OA+OB+OC>\frac{1}{2}(AB+BC+CA)$

Answer 1

In $\Delta ABC$,

$AB+AC>BC$ ….(i) [$\because$ Sum of two sides of a triangle is greater than the third side.]

Consider $\Delta OBC$,

$OB+OC>BC$ …(ii)

Subtracting (i) from (ii) we get,

$\Rightarrow (AB+AC)–(OB+OC)>(BC–BC)$

(1) $AB+AC>OB+OC$

From (1), $AB+AC>OB+OC$

Similarly, $AB+BC>OA+OC$

and $AC+BC>OA+OB$

Adding both sides of these three inequalities, we get,

$\Rightarrow (AB+AC)+(AB+BC)+(AC+BC)>(OB+OC)+(OA+OC)+(OA+OB)$

$\Rightarrow 2(AB+BC+AC)>2(OA+OB+OC)$

(2) $AB+BC+OA>OA+OB+OC$

Again in $\Delta ABC$,

$\Rightarrow OA+OB>AB$

$\Rightarrow OB+OC>BC$

$\Rightarrow OA+OC>AC$

On adding the above equations,

$\Rightarrow 2(OA+OB+OC)>AB+BC+CA$

Hence,

(3) $OA+OB+OC>\frac{1}{2}(AB+BC+CA)$