Weitzenböck's Inequality

The applet below illustrates a geometric proof of Weitzenböck's Inequality.

For any triangle ABC with sides $a,$ $b,$ $c,$ and the area $S,$

$a^2 + b^2 + c^2 \ge 4\sqrt{3}S.$

This wonderful geometric proof is due to C. Alsina and R. B. Nelsen.

23 November 2015, Created with GeoGebra

Explanation

References

C. Alsina and R. B. Nelsen, Geometric Proofs of Weitzenböck and Hadwiger-Finsler Inequalities, Mathematics Magazine, v. 81, No. 3, June 2008, pp. 216-219
C. Alsina, R. B. Nelsen, Charming Proofs, MAA, 2010, pp. 96-98

Theorem

For any $\Delta ABC$ with sides $a,$ $b,$ $c,$ and the area $S,$

(1)

$a^2 + b^2 + c^2 \ge 4\sqrt{3}S.$

Proof

Form, on the sides of $\Delta ABC,$ Napoleon triangles, $ABC',$ $BCA',$ $CAB'.$ For a triangle with all angles less than $120^{\circ},$ the three lines $AA',$ $BB',$ $CC'$ concur at Fermat's point of the triangle. The point solves the minimization problem: for all points $X,$ $XA + XB + XC$ attends its minimum at $F.$ If one of the angles is greater or equal to $120^{\circ}$ Fermat's point lies at the corresponding vertex.

Multiply (1) by $\sqrt{3}/4.$ The inequality becomes

(2)

$S_{a} + S_{b} + S_{c} \ge 3S,$

where S_{a} is the area of $\Delta BCA',$ etc.

Witzenbrock's inequality, PWW

In fact, a stronger result holds. Let $F$ be interior to $\Delta ABC,$ with distances $x,$ $y,$ $z$ to $A,$ $B,$ $C,$ respectively. Let $S_{k}$ denote the area of an equilateral triangle with side $k.$ Then

(3)

$S_{a} + S_{b} + S_{c} = 3S + S_{|x - y|} + S_{|y - z|} + S_{|z - x|}.$

To see that (3) holds, observe that $F$ splits $\Delta ABC$ into three triangles each with angle $120^{\circ}$ at $F.$ This shows that the remaining angles in each triangle add up to $60^{\circ}.$ It follows that three copies of each of the triangles fit into the corresponding equilateral triangle leaving equilateral triangles with sides $|x - y|,$ $|y - z|,$ and $|z - x|.$

When F coincides with, say, vertex C, then

(4)

$S_{a} + S_{b} + S_{c} \ge S_{c} \ge 3S.$

Stronger than (1) is the Hadwiger-Finsler inequality: if $a,$ $b,$ $c$ are the sides of a triangle with area $S$ then

(5)

$a^2 + b^2 + c^2 \ge 4\sqrt{3}S + (a - b)^2 + (b - c)^2 + (c - a)^2.$

In terms of areas, the latter is equivalent to

(6)

$S_{a} + S_{b} + S_{c} \ge 3S + S_{|a - b|} + S_{|b - c|} + S_{|c - a|}.$

In case $F$ is internal to $\Delta ABC,$ this inequality is a consequence of (3) because

$|x - y| \ge |a - b|,$ etc.

To prove these, assume $a \ge b \ge c.$ Then by reflecting the triangle with sides $b,$ $x,$ $z$ in the segment $z$ leads to the following configuration:

Hadwiger-Finsler inequality

The triangle inequality (in the yellow triangle) then implies $y - x \ge a - b.$ The other two inequalities are established similarly. Thus from (3)

$\begin{align} S_{a} + S_{b} + S_{c} &= 3S + S_{|x - y|} + S_{|y - z|} + S_{|z - a|}\\ &\ge 3S + S_{|a - b|} + S_{|b - c|} + S_{|c - a|}. \end{align}$

In case where, say, angle $C$ is $120^{\circ}$ or more so that $F$ coincides with $C,$ $z = 0,$ $x = b$ and $y = a.$ The inequality in (4) is refined to

$S_{a} + S_{b} + S_{c} \ge 3S + S_{|a - b|} + S_{a} + S_{b}.$

Now observe that $a \ge |b - c|$ and $b \ge |c - a|$ from which, again, (6) follows.

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