PlanetMath: proof of Hadwiger-Finsler inequality

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proof of Hadwiger-Finsler inequality

(Proof)

From the cosines law we get:

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

$\alpha$ being the angle between

and

. This can be transformed into:

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

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

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

Now remember that

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

and

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

Using this we get:

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

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

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

$\beta$ 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:

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

Using this we get:

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

This is the Hadwiger-Finsler inequality. $\Box$

"proof of Hadwiger-Finsler inequality" is owned by mathwizard.

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Cross-references: Hadwiger-Finsler inequality, convex, function, triangle, sides, angle, cosines law

This is version 2 of proof of Hadwiger-Finsler inequality, born on 2002-06-06, modified 2002-06-06.
Object id is 3062, canonical name is ProofOfHadwigerFinslerInequality.
Accessed 2020 times total.

Classification:

51M16 (Geometry :: Real and complex geometry :: Inequalities and extremum problems)

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