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Question Number 98983 by M±th+et+s last updated on 17/Jun/20
leta,b,cbepositiverealnumberssuchthatab+bc+ac=3provetheinqualitya(b2+c2)a2+bc+b(c2+a2)b2+ac+c(b2+a2)c2+ab⩾3
Commented by MJS last updated on 17/Jun/20
duetosymmetryextremesata=b=c⇒ab+bc+ac=3⇔3a2=3⇒a=±1buta>0⇒a=b=c=1theinequationwitha=b=cturnsinto3a⩾3⇒truefora=1nowtestifthisisminormaxbyputtinga=.999;b=1.001⇒c=1.0000005⇒lhs>3⇒provenIknowyouwantadifferentkindofproofbutthisistheeasiestpath
Commented by M±th+et+s last updated on 17/Jun/20
thisisagoodproofthankyou
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