câu 1:
a,b,c >0 thỏa ab+bc+ac=1
cmr: a^2/(a+bc) +b^2/(b+ac) +c^2/(c+ab) >=(a+b+c)/4
câu 2:
a,b,c>0 thỏa abc=1
cmr:
1/(a^3*(b+c))+1/(b^3*(a+c))+1/(c^3*(a+b))>=3/2
câu 2:
$\eqalign{
& S = {1 \over {{a^3}\left( {b + c} \right)}} + {1 \over {{b^3}\left( {c + a} \right)}} + {1 \over {{c^3}\left( {a + b} \right)}} \ge {3 \over 2} \cr
& quy\;dong: \cr
& S = {{ab + bc + ca + {a^4}{b^4} + {b^4}{c^4} + {c^4}{a^4} + {a^3}\left( {{b^2} + {c^2}} \right) + {b^3}\left( {{c^2} + {a^2}} \right) + {c^3}\left( {{a^2} + {b^2}} \right)} \over {\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}} \cr
& \cos i \cr
& ab + 2{a^3}{c^2} = ab + {a^3}{c^2} + {a^3}{c^2} \ge 3\root 4 \of {{a^7}b{c^4}} = 3{a^2}c\root 3 \of {abc} = 3{a^2}c \cr
& ac + 2{a^3}{b^2} \ge 3{a^2}b \cr
& ab + 2{b^3}{c^2} \ge 3{b^2}c \cr
& bc + 2{b^3}{a^2} \ge 3{b^2}a \cr
& ac + 2{c^3}{b^2} \ge 3{c^2}b \cr
& bc + 2{c^3}{a^2} \ge 3{c^2}a \cr
& 2\left( {{a^4}{b^4} + {b^4}{c^4} + {c^4}{a^4}} \right) \ge 6\root 3 \of {{a^8}{b^8}{c^8}} = 6 = 6abc \cr
& \to 2\left( {ab + bc + ca} \right) + 2\left( {{a^3}\left( {{b^2} + {c^2}} \right) + {b^3}\left( {{c^2} + {a^2}} \right) + {c^3}\left( {{a^2} + {b^2}} \right)} \right) + 2\left( {{a^4}{b^4} + {b^4}{c^4} + {c^4}{a^4}} \right) \ge 3\left( {{a^2}\left( {b + c} \right) + {b^2}\left( {c + a} \right) + {c^2}\left( {a + b} \right)} \right) + 6abc = 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) \cr
& \to S \ge {{3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)} \over {2\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}} = {3 \over 2} \cr
& dau = \leftrightarrow a = b = c = 1 \cr} $
câu 1 đề sai rồi xem lại đi bạn
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