3+1a+1b+1c=12(1a2+1b2+1c2)≥4(1a+1b+1c)2Đặt 1a+1b+1c=x(x>0)
Có 3+x≥4x2⇔4x2−x−3≤0⇔−34≤x≤1
Mà x>0⇒0<x≤1
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VT=1a+a+a+a+b+c+1b+b+b+b+c+a+1c+c+c+c+a+b
≤136(1a+1a+1a+1a+1b+1c)+136(4b+1c+1a)+136(4c+1a+1b)
=16x≤16
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Dấu "=" khi a=b=c=3