Question

chứng minh (1-a)(1-b)(1-c)>=8abc với a,b,c>=0 và a+b+c=1

Answer 1

Áp dụng BĐT AM-GM ta có:

\(VT=\left(1-a\right)\left(1-b\right)\left(1-c\right)\)

\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

\(\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ac}\)

\(=8abc=VP\)

Khi \(a=b=c\)

Answer 2

BĐT\(\Leftrightarrow\)(a+b)+(b+c)+(c+a)\(\ge\)8abc

TA có BDT cô si

a+b\(\ge\)2\(\sqrt{ab}\)

\(\Rightarrow\)(a+b)(b+c)(a+c)\(\ge\)\(2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ac}\)

Vậy (1-a)(1-b)(1-c)\(\ge\)8abc