Proof of a geometric statement

Question

If $D$ is a point inside a triangle $\triangle ABC$ then how the following statement is true.

statement: $AB+AC>BD+DC$.

I have tried in the following way but it seems to me defective.

$$\begin{align} AB+AC&>BC\tag{1}\\ BD+DC&>BC\tag{2} \end{align}$$

By subtracting $(1)-(2)$ we get,

$$AB+AC-BD-DC>0$$

or $$AB+AC>BD+DC$$

But if I subtract $(2)-(1)$ we get,

$$BD+DC>AB+AC$$

That is the problem. Can anyone help me to prove the above geometric statement?

Answer 1

Hint: Let $E$ and $F$ be points on $AB$ and $AC$ respectively such that $DE \parallel AC$ and $DF\parallel BA$.

Answer 2

The two suggestions are not helping since the greater sum segments may each be smaller than one of the segments in the smaller sum. The same way it is quite easy to create a case in which angel DAB is acute. Here is a solution: