Let a, B, C be the three sides of △ ABC, and s be the area of triangle. Prove that C ^ 2-A ^ 2-B ^ 2 + 4AB ≥ (4 √ 3) s

Let a, B, C be the three sides of △ ABC, and s be the area of triangle. Prove that C ^ 2-A ^ 2-B ^ 2 + 4AB ≥ (4 √ 3) s

We know that the triangle has three sides a, B, C, and the half circumference P = (a + B + C) / 2,
Then s = √ [P (P-A) (P-B) (P-C)] (Helen formula)
=√[(a+b+c)/2((-a+b+c)/2)((a-b+c)/2)((a+b-c)/2)]
=√[(a+b+c)(-a+b+c)(a-b+c)(a+b-c)]/4
=√[((a+b)^2-c^2)(c^2-(a-b)^2)]/4
Then (4 √ 3) s = √ [3 ((a + b) ^ 2-C ^ 2) (C ^ 2 - (a-b) ^ 2)]
If you cannot see clearly, you can make x = C ^ 2 - (a-b) ^ 2 > 0, y = (a + b) ^ 2-C ^ 2 > 0
(4√3)S=√(3xy)
x+y=4ab
x-y=2(c^2-a^2-b^2)
c^2-a^2-b^2+4ab=(x-y)/2+(x+y)=(3x+y)/2≥2√(3xy)/2=(4√3)S