The lengths of the sides of the triangle form an arithmetic progression

The lengths of the sides of the triangle form an arithmetic progression with a difference that is not zero. prove that only one angle of this triangle is greater than 60 °

1. Let:

a ≤ b ≤ c – sides of the triangle.
α ≤ β ≤ γ – corresponding angles.
Then:

a = b – d;
c = b + d, where
d> 0 – the difference of the arithmetic progression.
2. By the cosine theorem we have:

b ^ 2 = a ^ 2 + c ^ 2 – 2accosβ;
2accosβ = a ^ 2 + c ^ 2 – b ^ 2;
cosβ = 1/2 * (a ^ 2 + c ^ 2 – b ^ 2) / ac;
cosβ = 1/2 * ((b – d) ^ 2 + (b + d) ^ 2 – b ^ 2) / (b – d) (b + d);
cosβ = 1/2 * (b ^ 2 + 2d ^ 2) / (b ^ 2 – d ^ 2)> 1/2 * b ^ 2 / b ^ 2 = 1/2;
cosβ> 1/2;
β <60 °.
3. The average angle of a triangle is less than 60 °, which means that only one angle of this triangle is greater than 60 °. Which was required.