Prove inequalities in geometry

Question

Outline:

In a triangle $\triangle ABC$ let $\overline{AB}$ be the longest of the three sides. Let $G$ be the centroid of $\triangle ABC$ and $M$ the midpoint of $\overline{AB}$. Furthermore, let a point $D$ outside the triangle be given. The following inequalities now apply:

1. The distance from $D$ to the vertices of the triangle is less than $1$.
2. The side $\overline{AB}$ is bigger than $\sqrt{\frac{2}{3}}$ but less than $\sqrt 2$.
3. The median $\overline{CM}$ is bigger than $\frac{\sqrt 2}{2}$.

Prove: $\overline{DG}<\frac{\sqrt 2}{2}$.

I experimented a bit with Geogebra and found out by measuring side lengths that the statement holds. I can't find much on the internet about these kinds of problems with geometric inequalities. Therefore, I would be interested in what possibilities there are to approach such problems. I have tried to make progress with the triangle inequality and Ptolomew's inequality, but have not yet reached my goal, because the estimates were always too inaccurate.
Edit: The best possible bound to prove is $DG<\frac{\sqrt 7}{3}$.

Answer 1

I am getting a math contest practice problem vibe, so I will give you a couple of hints. Let me know if you want more details.

1. Prove $$DA^2+DB^2+DC^2 = 3DG^2 + GA^2+GB^2+GC^2.$$
2. Prove $$CM^2 = \dfrac{9}{4}GC^2 = \dfrac{2(AC^2+CB^2)-AB^2}{4}$$
3. Prove $$GA^2+GB^2+GC^2 = \dfrac{AB^2+BC^2+ CA^2}{3}$$ but this is a very well-known identity.

Then, using the bounds specific to your problem, you should be able to get an upper bound on $DG.$