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. $\Box$

Title proof of Hadwiger-Finsler inequality ProofOfHadwigerFinslerInequality 2013-03-22 12:45:21 2013-03-22 12:45:21 mathwizard (128) mathwizard (128) 5 mathwizard (128) Proof msc 51M16