三角形不等式在三角形ABC中,证明:(sinA+1/sinA)(
在三角形AB中, 证明:(sinA+1/sinA)(sinB+sinB)(sinC+1/sinC)>=(7根3)/6.
y=ln(sinx+1/sinx)=ln(1+sin^2x)-lnsinx 可以求二阶导数得y''≥0 所以∑ln(sinA+1/sinA)≥3ln[sin(A+B+C)/3+1/sin(A+B+C)/3] =3ln(7/(2√3)) 所以(sinA+1/sinA)(sinB+1/sinB)(sinC+1/sinC)≥343√3/72