proof of Hadwiger-Finsler inequality

From the cosines law we get:

$$a^2=b^2+c^2-2bc\cos \alpha ,$$

$\alpha$ being the angle between $b$ and $c$. This can be transformed into:

$$a^2={(b-c)}^2+2bc(1-\cos \alpha ).$$

Since $A=\frac{1}{2}bc\sin \alpha$ we have:

$$a^2={(b-c)}^2+4A\frac{1-\cos \alpha }{\sin \alpha }.$$

Now remember that

$$1-\cos \alpha =2{\sin }^2\frac{\alpha }{2}$$

and

$$\sin \alpha =2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}.$$

Using this we get:

$$a^2={(b-c)}^2+4A\tan \frac{\alpha }{2}.$$

Doing this for all sides of the triangle and adding up we get:

$$a^2+b^2+c^2={(a-b)}^2+{(b-c)}^2+{(c-a)}^2+4A\left(\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}+\tan \frac{\gamma }{2}\right).$$

$\beta$ and $\gamma$ being the other angles of the triangle. Now since the halves of the triangle’s angles are less than $\frac{\pi }{2}$ the function $\tan$ is convex we have:

$$\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}+\tan \frac{\gamma }{2}\ge 3\tan \frac{\alpha +\beta +\gamma }{6}=3\tan \frac{\pi }{6}=\sqrt{3}.$$

Using this we get:

$$a^2+b^2+c^2\ge {(a-b)}^2+{(b-c)}^2+{(c-a)}^2+4A\sqrt{3}.$$

This is the Hadwiger-Finsler inequality. $\mathrm{\square }$