From my school days in the 1960’s I remember using (and even proving) Heron’s formula for the area of a triangle, i.e. for ∆ABC the Area A = √(s(s-a)(s-b)(s-c)) where s is the semi-perimeter.

This formula is easy to remember, although I am not sure if s has any real significance other than to ‘simplify’ the formula and make it more memorable.

The formula can also be written as A² = p(a+b-c)(b+c-a)(c+a-b)/16, where p = a + b + c is the perimeter of the triangle.

The three terms (a + b – c), (b + c – a) and (c + a – b) are each > 0 and as such can be construed as the lengths of the edges of a cuboid. If so, then A²=pv/16 where v is the volume of the cuboid with edge lengths (a + b – c), (b + c – a) and (c + a – b).

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For a general cuboid of dimensions L, B and H, if we let P = 4(L + B + H), S = 2(LB + BH + LH) and V = LBH, and then apply the Arithmetic-Geometric mean inequality twice, it follows easily that … V≤(P/12)³ and V²≤(S/6)³, both of these with equality iff L = B = H. Interestingly if we combine the 2 inequalities above we get V≤PS/72. Also, the following equality holds, i.e. S²/4=L²B²+B²H²+L²H²+PV/2.

Now (LB-BH)²+(BH-LH)²+(LH-LB)²≥0, with equality iff L=B=H. From this it can be easily shown that L²B²+B²H²+L²H²≥VP/4 with equality iff L=B=H. Hence, using the equation S²/4=L²B²+B²H²+L²H²+PV/2, it follows that S²≥3PV, again with equality iff L=B=H.

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For ΔABC, if L = a + b – c, B = b + c – a, and H = c + a – b, then P = 4p and V≤(P/12)³ becomes v≤(p/3)³, with equality iff a = b = c. So, A²=pv/16 leads to the inequality A²≤p⁴/432 with equality iff ΔABC is equilateral.

This may also be written as A≤p²/(12√ 3) with equality iff ΔABC is equilateral. (*)

While working on this some time ago, I came across references to the Weitzenbock inequality on various websites, i.e. for ΔABC, the Area A≤(a²+b²+c²)/(4√ 3) with equality iff ΔABC is equilateral.

It is easy to show that for ΔABC, if p=a+b+c, p²≤3(a²+b²+c²),with equality iff a = b = c, and so Weitzenbock’s inequality can be derived from (*).

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If we now apply the other cuboid inequality V²≤(S/6)³ to the cuboid derived from the triangle, with edge lengths (a + b – c), (b + c – a) and (c + a – b), we get S=2((a+b-c)(a+c-b)+(a+c-b)(b+c-a)+(b+c-a)(a+b-c)) and this simplifies to S=2(p²-2(a²+b²+c²)).

Hence, V²≤(S/6)³ becomes v²≤(p² – 2(a² + b² + c²))³/27 which leads on to …

A⁴≤ p²(p² – 2(a² + b² + c²))³/(16²3³), with equality iff ΔABC is equilateral. (**)

Now p²≤3(a² + b² + c²) is equivalent to p² – 2(a² + b² + c²)≤p²/3 and so the above inequality for A ⁴ leads back naturally to (*).

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Now returning to A² = pv/16, and using S²≥3PV and P=4p, we get A≤S/(8√3) which is easily checked to be an equivalent form of the Hadwiger-Finsler inequality, again with equality iff ΔABC is equilateral.

It is straightforward to show that …

Hadwiger-Finsler => ** => * => Weitzenbock.

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If we start again with A²= pv/16 or (4A)²=pv and cube each side, we get (4A)⁶=p³v³.

Now p³≥27v and so (4A)⁶≥27v⁴ or (4A)⁶≥27(a+b-c)⁴(b+c-a)⁴(c+a-b)⁴ (***)

with equality iff a=b=c.

This is not dissimilar to the form of the Ono inequality,

i.e. (4A)⁶≥27(a²+b²-c²)²(b²+c²-a²)²(c²+a²-b²)².

Assuming that the triangle (a,b,c) is acute angled, the Ono inequality can be derived from (***) by examining the 3 ratios …

(a+b-c)²(b+c-a)²/((a²+b²-c²)(b²+c²-a²))

(a+c-b)²(b+c-a)²/((a²+c²-b²)(b²+c²-a²))

and

(a+b-c)²(a+c-b)²/((a²+b²-c²)(a²+c²-b²))

Each of these ratios can be shown easily to be ≥1 and from these inequalities the Ono inequality follows.

Suppose the first of these is <1.

Then (b²-(a-c)²)²/(b⁴-(a²-c²)²)<1.

It follows easily from the above that …

(a-c)²(a²+b²-c²)<0 and since (a,b,c) is acute angled this would imply that (a-c)²<0 which is clearly false. Hence the result follows for this ratio. The others follow using exactly the same argument. So,the Ono inequality is true if (a,b,c) is acute angled.

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If we introduce the variable m = p/3, the mean length of side of ΔABC, then (*) becomes A≤√3(m²/4) where √3(m²/4) is the area of the equilateral triangle of side m.

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If ΔABC is right-angled at C, i.e. C = 90º, then if p(perimeter) = a + b + c,

(p – c)² = (a + b)² = a² + 2ab + b² = c² + 4(Area of Δ).  [Since Area of Δ = 1/2 ab]

So, p² – 2pc = 4(Area of Δ) and so c = p/2 – 2(Area of Δ)/(p).

For a non-equilateral triangle, we have that Area of Δ < p²/(12√3) and so it follows that

c > p/2 – p/(6√3) = p(3√3 -1)/(6√3).  It is straightforward to show also that c < p/2.

So, p/2 > c > p/2 – p/(6√3) = p(3√3 -1)/(6√3).    [(3√3 -1)/(6√3) = 0.4037749551]

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Extra … TBC

The area of a triangle satisfies the inequality A² ≤ p⁴/432 where p is the perimeter. So, for the triangle (X,Y,Z) formed from the corresponding angle sizes of the acute triangle (x,y,z) we have (using Heron’s formula) and the above inequality that

π(X + Y – Z)(X + Z – Y)(Y + Z – X)/16 ≤ π⁴/432.

Using the fact that X+Y+Z = π, this can be simplified to

(π/2 – X)(π/2 – Y)(π/2 – Z)≤π³ /216, clearly with equality iff (x,y,z) is equilateral.

This last inequality can also be proved directly by applying the AGM inequality to the angle sizes π/2 – X, π/2 – Y and π/2 – Z assuming that each is >0.

In the same way it can also be shown that XYZ≤π³/27 for the angles X, Y and Z in any triangle, with equality iff X = Y = Z.

If (x, y, z) is an acute angled triangle, with sides x<=y<=z and corresponding angles X, Y and Z , then the inequalities involving the angles sizes, i.e. π/2 – X < 2YZ/π, π/2 – Y < 2XZ/π and

π/2 – Z ≤ 3XY/2π, become (using XYZ≤π³/27)  …

π/2 – X < 2π²/27X, π/2 – Y < 2π²/27Y and  π/2 – Z ≤ π²/18Z (the last with equality iff Z = π/3).

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