>>956 (1)　コーシーで
　(1+1+1)^5 (a^6 + b^6 + c^6) ≧ (a+b+c)^6,
一方、
　a = r {cot(B/2) + cot(C/2)},
　b = r {cot(C/2) + cot(A/2)},
　c = r {cot(A/2) + cot(B/2)},
∴　a+b+c = 2r {cot(A/2) + cot(B/2) + cot(C/2)} ≧ 6r cot((A+B+C)/6) = 6r cot(π/6) =（6√3）r