Let's say we have $\triangle ABC$ and inside it the point $D$. Prove that $AB+BC>AD+DC$.
Well, I don't know if it's 100% true all the time and that's why I want to prove it. In other words, it's a small hypothesis of mine. Anyone can dis/prove it?
My try :
It seems it is quite possible to prove it by using the cosine rule(by making squares). However, I would like a solution using simple things such as triangle inequalities and the rule of big angle-big side. On the second thought, I tried goin a bit far with the cosine rule: Let's denote $AB, BC, AD, DC$ as $a, b, c, d$ respectively. $\angle ABC= \alpha$, $\angle ADC = \beta$.
Then, $a^2+b^2-2ab \cos\alpha =c^2 +d^2-2cd\cos\beta$
$(a+b) ^2-(c+d)^2=2ab(1+\cos\alpha) - 2cd(1+\cos\beta) $,
Now, all we have to prove is that the right side is greater than zero. We know $\alpha>\beta$, but the above statement is still not easily provable.