Triangle inequality for an obtuse triangle

Question

$\alpha < 45^\circ$, how to show that

1) $|AB+AC|>|DB+DC|$?

2) $|AB+AC|>|DB+DC+DA|$?

Answer 1

For 1), note that the point $A$ is on an ellipse with foci at $B,C$ with length sum $|AB|+|BC|.$ The equality of the two marked angles $\alpha$ means that the ray $\vec{AD}$ is perpendicular to the ellipse at $A$, pointing into the ellipse. If as suggested by the diagram the point $D$ is in fact interior to the triangle $ABC$, then it follows that $D$ is interior to the ellipse, so that the length sum $|DB|+|DC|<|AB|+|BC|.$

Note that this argument didn't use $\alpha <45^{\circ}.$ Do you know of a case where inequality 1) fails when $\alpha$ is more than 45 degrees?

I haven't come up with anything (yet) on inequality (2).

NOTE on further looking at it, you don't need equality of the angles marked $\alpha$ in the diagram, or any inequality on the two angles there. All you need is that the point $D$ is interior to the triangle $ABC$, which the diagram clearly intends. Given any ellipse with foci at $B,C$, and any point $P$ on that ellipse, the triangle $BCP$ lies entirely in the interior of the ellipse, except for the point $P$ itself. It follows that any point $Q$ interior to triangle $BCP$ (so other than $P$) will have smaller length sum to the foci $B,C$ than does the point $P$ on the ellipse.

Inequality 2) counterexample:

Let $A=(0,0),B=(-1,9/10),C=(1,9/10),D=(0,8/10).$

Then $\alpha=\arctan(9/10) \approx 41.98^\circ,$ so that fits, and it's easy to see that $D$ is interior to triangle $ABC$. But computing AB+AC gives $\sqrt{181}/5 \approx 2.69072$, while computing the right side $DB+DC+DA$ gives $1/8+\sqrt{101}/5 \approx 2.80998,$ so that in this example the inequality stated in the OP as 2) turns out to have the right side greater than the left. There are also cases where the inequality does hold, e.g. if $D$ above is replaced by $(0,1/2)$. This shows there isn't an inequality either way between these two quantities, under the given assumptions.