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 A^{2} = 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 A^{2}=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 V^{2}≤(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. S^{2}/4=L^{2}B^{2}+B^{2}H^{2}+L^{2}H^{2}+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 L^{2}B^{2}+B^{2}H^{2}+L^{2}H^{2}≥VP/4 with equality iff L=B=H. Hence, using the equation S^{2}/4=L^{2}B^{2}+B^{2}H^{2}+L^{2}H^{2}+PV/2, it follows that S^{2}≥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, A^{2}=pv/16 leads to the inequality A^{2}≤p^{4}/432 with equality iff ΔABC is equilateral.

This may also be written as A≤p^{2}/(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≤(a^{2}+b^{2}+c^{2})/(4√ 3) with equality iff ΔABC is equilateral.

It is easy to show that for ΔABC, if p=a+b+c, p^{2}≤3(a^{2}+b^{2}+c^{2}),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 V^{2}≤(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(p^{2}-2(a^{2}+b^{2}+c^{2})).

Hence, V^{2}≤(S/6)^{3} becomes v^{2}≤(p^{2} – 2(a^{2} + b^{2} + c^{2}))^{3}/27 which leads on to …

A^{4}≤ p^{2}(p^{2} – 2(a^{2} + b^{2} + c^{2}))^{3}/(16^{2}3^{3}), with equality iff ΔABC is equilateral. (**)

Now p^{2}≤3(a^{2} + b^{2} + c^{2}) is equivalent to p^{2} – 2(a^{2} + b^{2} + c^{2})≤p^{2}/3 and so the above inequality for A ^{4} leads back naturally to (*).

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Now returning to A^{2} = pv/16, and using S^{2}≥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 A^{2}= pv/16 or (4A)^{2}=pv and cube each side, we get (4A)^{6}=p^{3}v^{3}.

Now p^{3}≥27v and so (4A)^{6}≥27v^{4 }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(a^{2}+b^{2}-c^{2})^{2}(b^{2}+c^{2}-a^{2})^{2}(c^{2}+a^{2}-b^{2})^{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}/((a^{2}+b^{2}-c^{2})(b^{2}+c^{2}-a^{2}))

(a+c-b)^{2}(b+c-a)^{2}/((a^{2}+c^{2}-b^{2})(b^{2}+c^{2}-a^{2}))

and

(a+b-c)^{2}(a+c-b)^{2}/((a^{2}+b^{2}-c^{2})(a^{2}+c^{2}-b^{2}))

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 (b^{2}-(a-c)^{2})^{2}/(b^{4}-(a^{2}-c^{2})^{2})<1.

It follows easily from the above that …

(a-c)^{2}(a^{2}+b^{2}-c^{2})<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(m^{2}/4) where √3(m^{2}/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 A^{2} ≤ p^{4}/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 ≤ π^{4}/432.

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

(π/2 –** X**)(π/2 – **Y**)(π/2 – **Z**)≤π^{3} /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** < 2**YZ**/π, π/2 –** Y** < 2**XZ**/π and

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

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

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