Ta có: \(\left(a+b+c\right)^3-27abc=\frac{7a+b+c}{2}\left(b-c\right)^2+\frac{7b+c+a}{2}\left(c-a\right)^2+\frac{7c+a+b}{2}\left(a-b\right)^2\)
Va` \(a+b+c-\sqrt{3\left(a^2+b^2+c^2\right)}=\frac{-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2}{a+b+c+\sqrt{3\left(a^2+b^2+c^2\right)}}\)
\(BDT\Leftrightarrow\frac{\left(a+b+c\right)^3}{abc}-27+54\left(\frac{a+b+c}{\sqrt{3\left(a^2+b^2+c^2\right)}}-1\right)\ge0\)
\(\Leftrightarrow\frac{\left(a+b+c\right)^3-27abc}{abc}+54\left(\frac{a+b+c-\sqrt{3\left(a^2+b^2+c^2\right)}}{\sqrt{3\left(a^2+b^2+c^2\right)}}\right)\ge0\)
\(\Leftrightarrow\frac{Σ\left(\frac{7c+a+b}{2}\left(a-b\right)^2\right)}{abc}-\frac{\frac{Σ54\left(a-b\right)^2}{a+b+c+\sqrt{3\left(a^2+b^2+c^2\right)}}}{\sqrt{3\left(a^2+b^2+c^2\right)}}\ge0\)
\(\LeftrightarrowΣ\left(a-b\right)^2\left(\frac{\frac{7c+a+b}{2}}{abc}-\frac{\frac{54}{a+b+c+\sqrt{3\left(a^2+b^2+c^2\right)}}}{\sqrt{3\left(a^2+b^2+c^2\right)}}\right)\ge0\) *Đúng*
"=" <=> a=b=c :v
BĐT \(\Leftrightarrow\frac{\left(a+b+c\right)^3}{abc}-27+54\left[\frac{\left(a+b+c\right)}{\sqrt{3\left(a^2+b^2+c^2\right)}}-1\right]\ge0\)
\(VT-VP=\left[\frac{27}{4bc}-\frac{\frac{3}{2\left[\Sigma_{cyc}a+\sqrt{3\left(\Sigma_{cyc}a^2\right)}\right]}}{\sqrt{3\left(\Sigma_{cyc}a^2\right)}}\right]\left(b-c\right)^2+\left[\frac{\left(a+4b+4c\right)}{4abc}-\frac{\frac{1}{2\left[\Sigma_{cyc}a+\sqrt{3\left(\Sigma_{cyc}a^2\right)}\right]}}{\sqrt{3\left(\Sigma_{cyc}a^2\right)}}\right]\left(b+c-2a\right)^2\)
Và bí:((Nhìn xấu quá ngán xử lí..
AM-GM: \(VT\ge\frac{\left(a+b+c\right)^2}{abc}+\frac{108}{\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}+\left(a+b+c\right)}\ge\frac{81}{a+b+c}\)
Đổi biến sang uvw là xong :D
