Question

ch a b c >0 va a+b+c=1 cmr

\(6\left(ab+bc+ac\right)+a\left(b-c\right)^2+b\left(a-c\right)^2+c\left(a-b\right)^2< =2\)

Answer 1

Ta cần chứng minh

\(6\left(a+b+c\right)\left(ab+bc+ac\right)+Σ\left(a^2b+a^2c-2abc\right)\le2\left(a+b+c\right)^3\)

\(\Leftrightarrow6Σ\left(a^2b+a^2c+abc\right)+Σ\left(a^2b+a^2c-2abc\right)\le2Σ\left(a^3+3a^2b+3a^2c+2abc\right)\)

\(\LeftrightarrowΣ\left(2a^3-a^2b-a^2c\right)\ge0\)

\(\LeftrightarrowΣ\left(a^3-a^2b-ab^2+b^3\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)+\left(b-c\right)^2\left(b+c\right)+\left(c-a\right)^2\left(c+a\right)\ge0\)