Theo BĐT AM-GM ta có:
\(x^4+x^4+x^4+\left(\dfrac{2}{3}\right)^4\ge4\sqrt[4]{x^4x^4x^4.\left(\dfrac{2}{3}\right)^4}=\dfrac{8}{3}x^3\left(1\right)\)
Tương tự: \(y^4+y^4+y^4+\left(\dfrac{2}{3}\right)^4\ge\dfrac{8}{3}y^3\left(2\right)\); \(z^4+z^4+z^4+\left(\dfrac{2}{3}\right)^4\ge\dfrac{8}{3}z^3\left(3\right)\)
Cộng theo vế (1), (2), (3) ta có:
\(3\left(x^4+y^4+z^4\right)+\dfrac{16}{27}\ge\dfrac{8}{3}\left(x^3+y^3+z^3\right)\)
\(\Leftrightarrow x^4+y^4+z^4\ge\dfrac{8}{9}\left(x^3+y^3+z^3\right)-\dfrac{16}{81}\)
Ta có: \(S=1+\dfrac{1}{2}\left(x^4+y^4+z^4\right)-x^3-y^3-z^3\ge1+\dfrac{1}{2}\left[\dfrac{8}{9}\left(x^3+y^3+z^3\right)-\dfrac{16}{81}\right]-x^3-y^3-z^3=\dfrac{73}{81}-\dfrac{5}{9}\left(x^3+y^3+z^3\right)\)
