Inequality (b)/(ab+b+1)+(c)/(bc+c+1)+(a)/(ac+a+1)>=(3m)/(m^2+m+1) Let m=(abc)^(1/3), where a,b,c in RR^+. In this inequality I first applied Titu's lemma ; then Rhs will come 9/(some terms) ; now to maximise the rhs I tried to minimise the denominator by applying AM-GM inequality.But then the reverse inequality is coming Please help.

Orion Cervantes 2022-09-25 Answered
Inequality b a b + b + 1 + c b c + c + 1 + a a c + a + 1 3 m m 2 + m + 1
Let m = ( a b c ) 1 3 , where a , b , c R + . Then prove that
b a b + b + 1 + c b c + c + 1 + a a c + a + 1 3 m m 2 + m + 1
In this inequality I first applied Titu's lemma ; then Rhs will come 9/(some terms) ; now to maximise the rhs I tried to minimise the denominator by applying AM-GM inequality.But then the reverse inequality is coming Please help.
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Answers (1)

edytorialkp
Answered 2022-09-26 Author has 10 answers
By Holder:
c y c b a b + b + 1 = 1 ( a b c 1 ) 2 c y c ( a b + b + 1 ) 1 ( m 3 1 ) 2 ( a 2 b 2 c 2 3 + a b c 3 + 1 ) 3 =
= 1 ( m 1 ) 2 ( m 2 + m + 1 ) 2 ( m 2 + m + 1 ) 3 = 1 ( m 1 ) 2 m 2 + m + 1 = 3 m m 2 + m + 1 .
Done!
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