∵3sin2B+7sin2C=2sinAsinBsinC+2sin2A,由正弦定理可得:3b2+7c2=2bcsinA+2a2,∴a2=3b2+7c2-2bcsinA2,又a2=b2+c2-2bccosA,∴3b2+7c2-2bcsinA2=b2+c2-2bccosA,化为:2(sinA-2cosA)=b2+5c2bc=bc+5cb≥25,当且...