Triangle inequalities.

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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).


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.


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 (*).


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 (*).


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.


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 …





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.


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.


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]


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


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