# Triangle inequalities.

· Uncategorized
Authors

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 …

(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.

————————————————-

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).

——————————————-