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: $$\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}+\tan\frac{\gamma}{2} \geq 3\tan\frac{\alpha+\beta+\gamma}{6} =3\tan\frac{\pi}{6}=\sqrt{3}.$$ Using this we get: $$a^2+b^2+c^2\geq (a-b)^2+(b-c)^2+(c-a)^2+4A\sqrt{3}.$$ This is the Hadwiger-Finsler inequality.