在三角形ABC中,角A、B、C的对边分别为a、b、c,且A+C≤2B.(1)求证：B≥π/3；(2)求证：a+c≤2b.

1) A+B+C = π
A+C = π-B
A+C≤2B
π-B≤2B
3B≥π
B≥π/3
2)由正弦定理
a/sinA=b/sinB=C/sinC=2R
a = 2R*sinA
b = 2R*sinB
c = 2R*sinC
2b = 4R*sinB = 4R*sin[π-(A+C)] = 4R*[sinπ*cos(A+C)-cosπ*sin(A+C)]
= 4R*sin(A+C) = 4R*[sinA*cosC+cosA*sinC)
a+c = 2R*(sinA+sinC)
...正余弦定理反复代入吧...