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'proof of Hadwiger-Finsler inequality'
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| Title of object: |
proof of Hadwiger-Finsler inequality |
| Canonical Name: |
ProofOfHadwigerFinslerInequality |
| Type: |
Proof |
| Created on: |
2002-06-06 11:50:29-04 |
| Modified on: |
2002-06-06 11:50:29-04 |
| Classification: |
msc:51M16 |
Preamble:
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Content:
From the cosine's 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
$$1-\cos\alpha=2\sin^2\frac{\alpha}{2}$$
$$\sin\alpha=2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}.$$
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. $\Box$ |
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