The difference between any two sides of a triangle is than the third side. (greater/ smaller)

$ABCD$ is a quadrilateral. Prove that $(AB+BC+CD+DA)>(AC+BD)$

In the given figure, $P$ is a point on the side $BC$ of $∆ABC$. Prove that $(AB+BC+CA)>2AP$.

If $O$ is a point in the exterior of $∆ABC$, show that $2(OA+OB+OC)>(AB+BC+CA)$.

Which of the following statements is TRUE?

Statement $–1$ : The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

Statement $–\mathit{}2$ : If $P$ is a point on the side $BC$ of $ABC$. Then $(AB+BC+AC)>2AP$.

$ABCD$ is a quadrilateral. Is $AB\mathrm{}+\mathrm{}BC\mathrm{}+\mathrm{}CD\mathrm{}+\mathrm{}DA\mathrm{}<\mathrm{}2(AC\mathrm{}+\mathrm{}BD)$?

$AM$ is a median of $∆ABC$. Prove that $(AB+BC+CA)>2AM$.

Is it possible to have a triangle with the following sides?

$2 \mathrm{c}\mathrm{m},3 \mathrm{c}\mathrm{m},5 \mathrm{c}\mathrm{m}$