prop.'s of trianges

h₁ = b sin C = c sin B
h₂ = a sin C = c sin A
h₃ = a sin B = b sin A

a²/h₁² + b²/h₂² = a²/b² sin²C + b²/a² sin²C ≥ 2/sin²C from AM-GM Inequality

Likewise

b²/h₂² + c²/h₃² ≥ 2/sin²B and

c²/h₃² + a²/h₁² ≥ 2/sin²A

Hence a²/h₁² + b²/h₂² + c²/h₃² ≥ (1/sin²A + 1/sin²B + 1/sin²C)

But we know that

sin²A + sin²B + sin²C ≤ 9/4

From AM-GM, we have
(sin²A + sin²B + sin²C) (1/sin²A + 1/sin²B + 1/sin²C) ≥ 9

Hence, (1/sin²A + 1/sin²B + 1/sin²C) ≥ 4

From which we infer that a²/h₁² + b²/h₂² + c²/h₃² ≥ 4