Here is the details: First dividing 4 on both sides and moving 27(4 +β)2on the right hand side:4(6 +β)327(4 +β)2≥4⇐⇒(6 +β)3≥27(4 +β)2(1)Then we may do a subu= 4 +β, then we get(6 +β)3≥27(4 +β)2⇐⇒(u+ 2)3≥27u2(2)then expand (u+ 2)3=u3+ 6u2+ 12u+ 8, then we get(u+ 2)3≥27u2⇐⇒u3+ 6u2+ 12u+ 8≥27u2(3)Then move 27u2on the left hand sideu3+ 3u2+ 12u+ 8≥27u2⇐⇒u3-21u2+ 12u+ 8≥0(4)Then our goal is to solve the cubic inequality(we may just take =, our goal is to find the smallestβ), there is no formula for us to solve it directly. So we usually guess a solution for this, it turns outthatu= 1 is a solution, that means we find one factor ((u-1)) for the left hand side of (5)u3-21u2+ 12u+ 8 = 0(5)Here we may write downu3-21u2+ 12u+ 8 = (u-1)(au2+bu+c)(6)herea, b, care constants. You may expand right hand side of (7) and then compare the coef. of both
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