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Hadwiger–Finsler inequality

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Inequality applicable to triangles

In mathematics, the Hadwiger–Finsler inequality is a result on the geometry of triangles in the Euclidean plane. It states that if a triangle in the plane has side lengths a, b and c and area T, then

a 2 + b 2 + c 2 ≥ ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 + 4 3 T (HF) . {\displaystyle a^{2}+b^{2}+c^{2}\geq (a-b)^{2}+(b-c)^{2}+(c-a)^{2}+4{\sqrt {3}}T\quad {\mbox{(HF)}}.}

Related inequalities

a 2 + b 2 + c 2 ≥ 4 3 T (W) . {\displaystyle a^{2}+b^{2}+c^{2}\geq 4{\sqrt {3}}T\quad {\mbox{(W)}}.}

Hadwiger–Finsler inequality is actually equivalent to Weitzenböck's inequality. Applying (W) to the circummidarc triangle gives (HF)

Weitzenböck's inequality can also be proved using Heron's formula, by which route it can be seen that equality holds in (W) if and only if the triangle is an equilateral triangle, i.e. a = b = c.

  • A version for quadrilateral: Let ABCD be a convex quadrilateral with the lengths a, b, c, d and the area T then:
a 2 + b 2 + c 2 + d 2 ≥ 4 T + 3 − 1 3 ∑ ( a − b ) 2 {\displaystyle a^{2}+b^{2}+c^{2}+d^{2}\geq 4T+{\frac {{\sqrt {3}}-1}{\sqrt {3}}}\sum {(a-b)^{2}}} with equality only for a square.

Where ∑ ( a − b ) 2 = ( a − b ) 2 + ( a − c ) 2 + ( a − d ) 2 + ( b − c ) 2 + ( b − d ) 2 + ( c − d ) 2 {\displaystyle \sum {(a-b)^{2}}=(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}+(c-d)^{2}}

Proof

From the cosines law we have:

a 2 = b 2 + c 2 − 2 b c cos ⁡ α {\displaystyle a^{2}=b^{2}+c^{2}-2bc\cos \alpha }

α being the angle between b and c. This can be transformed into:

a 2 = ( b − c ) 2 + 2 b c ( 1 − cos ⁡ α ) {\displaystyle a^{2}=(b-c)^{2}+2bc(1-\cos \alpha )}

Since A=1/2bcsinα we have:

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

Now remember that

1 − cos ⁡ α = 2 sin 2 ⁡ α 2 {\displaystyle 1-\cos \alpha =2\sin ^{2}{\frac {\alpha }{2}}}

and

sin ⁡ α = 2 sin ⁡ α 2 cos ⁡ α 2 {\displaystyle \sin \alpha =2\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}}

Using this we get:

a 2 = ( b − c ) 2 + 4 A tan ⁡ α 2 {\displaystyle 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 + 4 A ( tan ⁡ α 2 + tan ⁡ β 2 + tan ⁡ γ 2 ) {\displaystyle a^{2}+b^{2}+c^{2}=(a-b)^{2}+(b-c)^{2}+(c-a)^{2}+4A(\tan {\frac {\alpha }{2}}+\tan {\frac {\beta }{2}}+\tan {\frac {\gamma }{2}})}

β and γ being the other angles of the triangle. Now since the halves of the triangle’s angles are less than π/2 the function tan is convex we have:

tan ⁡ α 2 + tan ⁡ β 2 + tan ⁡ γ 2 ≥ 3 tan ⁡ α + β + γ 6 = 3 tan ⁡ π 6 = 3 {\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:

a 2 + b 2 + c 2 ≥ ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 + 4 3 A {\displaystyle a^{2}+b^{2}+c^{2}\geq (a-b)^{2}+(b-c)^{2}+(c-a)^{2}+4{\sqrt {3}}\,A}

This is the Hadwiger-Finsler inequality.

History

The Hadwiger–Finsler inequality is named after Paul Finsler and Hugo Hadwiger (1937), who also published in the same paper the Finsler–Hadwiger theorem on a square derived from two other squares that share a vertex.

See also

References

  1. Martin Lukarevski, The circummidarc triangle and the Finsler-Hadwiger inequality, Math. Gaz. 104 (July 2020) pp. 335-338. doi:10.1017/mag.2020.63
  2. Leonard Mihai Giugiuc, Dao Thanh Oai and Kadir Altintas, An inequality related to the lengths and area of a convex quadrilateral, International Journal of Geometry, Vol. 7 (2018), No. 1, pp. 81 - 86,

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