Giả sử \(c=max\left\{a,b,c\right\}\)
BĐT \(\Leftrightarrow a^4+b^4+c^4\ge\frac{a+b+c}{3}\left(a^3+b^3+c^3\right)\)
\(\Leftrightarrow3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
\(VT-VP=\frac{1}{8}\left[\left(b+c-2a\right)^2\left\{3a^2+\left(a+b+c\right)^2\right\}+3\left(5b^2+6bc+5c^2-2ab-2ac\right)\right]\ge0\)