If (a,b,c) are pythagorean triples such as a^2+b^2=c^2 then prove a^3+b^3+3ab>c^3

Blocked on this for the past hour

I understand they must be positive numbers

Clearly c>a and c>b

but and thinking answer around with using (a+b)^3=a3 + 3a2b + 3ab2 + b3

or (a+b+c)^3 = a^3+b^3+c^3 + 3(a+b)(b+c)(a +c)

but can't find the trick. Anyone can help? Thanks!

Check your formula - it is incorrect. It works for the Pythagorean triplet 3,4,5:

3^3 + 4^3 + 3(3)(4) = 127;\ \ 5^3 = 125

but not for the triplet 5,12,13:

5^3 + 12^3 + 3(5)(12) = 2033; \ \ 13^3 = 2197

EDIT - I wonder if the correct formulation of this problem should be to prove that:

a^3 + b^3 + 3ab(a+b) \gt c^3 ?? That would be a breeze (and applies to ALL triangles, not just right angle triangles with integer sides).