Triangle Inequality?

It can be proved that for ΔABC,

cot A + cot B + cot C ≥ √3

=> (b^2 + c^2 - a^2) / 4S + (c^2 + a^2 - b^2) / 4S + (a^2 + b^2 - c^2) / 4S ≥ √3

=> a^2 + b^2 + c^2 >= (4S)√3.

Equality holds for a = b = c (because, then cotA + cotB + cot C = √3.)

cotA = (b^2 + c^2 - a^2) / 4S, etc. follow from the formulae of area of triangle which you may get in your text-book in the chapter on properties of triangle.

PROOF OF: cotA + cotB + cot C = √3

For ΔABC, A + B + C = π

=> tan(A + B) = tan(π - C)

=> (tan A + tan B) / (1 - tan A tan B) = - tan C

=> tan A + tan B + tan C = tan A tan B tan C

(dividing by tan A tan B tan C)

=> cot B cot C + cot C cot A + cot A cot B = 1 ... ( 1 )

Now, (cot A - cot B)^2 + (cot B - cot C)^2 + (cot C - cotA)^2 ≥ 0

=> 2(cot^2 A + cot^2 B + cot^2 C) - 2(cot A cot B + cot B cot C + cot C cot A) ≥ 0

[cancelling 2 and using the equation ( 1 )]

cot^2 A + cot^2 B + cot^2 C ≥ 1

=> cot^2 A + cot^2 B + cot^2 C + 2(cot A cot B + cot B cot C + cot C cot A) ≥ 3

[Again using equation ( 1 )]

=> (cot A + cot B + cot C)^2 ≥ 3

=> cot A + cot B + cot C ≥ √3

This is the so called Hadwiger's inequality /a Swiss mathematician/ and there are many ways to prove it. I'll suggest a proof, relating this problem to another interesting one - concerning Napoleon's triangles - please follow the link to the Wiki article below:

http://en.wikipedia.org/wiki/Napoleon_theorem

The 1st sentence in the article is incomplete, it should continue like this:

(***) ". . . the centroids . . , or COINCIDE (in the inward case)."

This theorem can be easily proven circumscribing 3 circles around the triangles ABZ, BCX and CAY on the picture. See also

http://en.wikipedia.org/wiki/Fermat_point

Now let us find the side lengths of both Napoleon's triangles. Let angle ACB = γ is the greatest angle in ABC (at least 60°),

R = abc/(4S) - radius of the circumscribed circle,

sinγ = c/(2R), cosγ = (a² + b² - c²)/(2ab); having

|CB| = a, |CA| = b, the triangle CLM yields:

|CL| = a/√3, |CM| = b/√3 (2/3 of the medians of an equilateral triangles), angle LCM = γ + 2*30°. Imagine the inner Napoleon triangle CL'M', the only difference will be angle L'CM' = γ - 2*30°, so its side length

|L'M'|² = |CL'|² + |CM'|² - 2 |CL'|*|CM'| cos(γ - 60°) =

= a²/3 + b²/3 - 2(ab/3)*(cosγ*cos60° + sinγ*sin60°) =

= a²/3 + b²/3 - 2(ab/3)*(a² + b² - c²)/(2ab)*(1/2) -

- 2(ab/3)*(c/(2R))√3/2 =

= (a² + b² + c²)/6 - abc√3/(6R) = (a² + b² + c² - 4S√3)/6,

/similarly |LM|² = (a² + b² + c² + 4S√3)/6/

and |L'M'|² ≥ 0 produces the desired result, equality here we have /look (***) above/ iff

L' ≡ M' ≡ N', the latter equivalent to a = b = c.

a^2 + b^2 + c^2 ≥ S (4sqrt 3)

c^2 = a^2 + b^2 - 2ab cos C

a^2 + b^2 + (a^2 + b^2 - 2ab cos C) ≥ S (4sqrt 3)

2a^2 + 2b^2 - 2ab cos C ≥ S (4sqrt 3)

a^2 + b^2 - ab cos C ≥ S (2sqrt 3)

c^2 ≥ S (2sqrt 3) - ab cos C

S = (1/2)(ab)(sin C)

c^2 ≥ (sqrt 3)(ab)(sin C) - (ab)(cos C)

c^2 ≥ ab [ (sqrt 3)(sin C) - cos C ]

c^2 ≥ ab [ (sqrt 3)(sin C) - sin (pi/2 - C) ]

(sqrt 3)(sin C) - sin (pi/2 - C) has a maximum of sqrt 3 and a minimum of -1

-1 ≤ (sqrt 3)(sin C) - sin (pi/2 - C) ≤ sqrt 3

-ab ≤ ab [ (sqrt 3)(sin C) - sin (pi/2 - C) ] ≤ ab sqrt 3

c^2 ≥ (ab)(sqrt 3) ≥ -ab

The only way ab sqrt 3 is the maximum, is if the triangle is right triangle.

a^2 + b^2 ≥ (ab)(sqrt 3)

To maximize ab sqrt 3, a = b

a^2 + a^2 ≥ a^2(sqrt 3)

2a^2 ≥ a^2 (sqrt 3)

2 ≥ sqrt 3 is true, therefore our statement is also true.

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EDIT:

I screwed up with the Pythagorean Theorem and maximizing part... I'll fix it IF I can.

use cosine laws a^2 = b^2 + c^2 - 2bc cos A etc & area formula area = (1/2)(base)(height).