a,b,c >0 ;a+2b+3c=14
tìm GTNN
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2 câu trả lời 244
A=Σ1a5(b+2c)2=Σ(abc)3a5(b+2c)2=Σb3c3a2(b+2c)2=Σb3c3(ab+2ac)2A=Σ1a5(b+2c)2=Σ(abc)3a5(b+2c)2=Σb3c3a2(b+2c)2=Σb3c3(ab+2ac)2
có:b3c3(ab+2ac)2+ab+2ac27+ab+2ac27≥33√b3c3272=bc3có:b3c3(ab+2ac)2+ab+2ac27+ab+2ac27≥3b3c32723=bc3
tươngtương tự⇒A≥(bc+ac+ab3)−(2(ab+2ac)+2(bc+2ab)+2(ca+2bc)27)tự⇒A≥(bc+ac+ab3)−(2(ab+2ac)+2(bc+2ab)+2(ca+2bc)27)
=9(ab+bc+ac)−2(ab+2ac+bc+2ab+ca+2bc)27=3(ab+bc+ca)27≥327.33√(abc)2=
A=Σ1a5(b+2c)2=Σ(abc)3a5(b+2c)2=Σb3c3a2(b+2c)2=Σb3c3(ab+2ac)2A=Σ1a5(b+2c)2=Σ(abc)3a5(b+2c)2=Σb3c3a2(b+2c)2=Σb3c3(ab+2ac)2
có:b3c3(ab+2ac)2+ab+2ac27+ab+2ac27≥33√b3c3272=bc3có:b3c3(ab+2ac)2+ab+2ac27+ab+2ac27≥3b3c32723=bc3
tươngtương tự⇒A≥(bc+ac+ab3)−(2(ab+2ac)+2(bc+2ab)+2(ca+2bc)27)tự⇒A≥(bc+ac+ab3)−(2(ab+2ac)+2(bc+2ab)+2(ca+2bc)27)
=9(ab+bc+ac)−2(ab+2ac+bc+2ab+ca+2bc)27=3(ab+bc+ca)27≥327.33√(abc)2=
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