Sửa đề \(\dfrac{\left(x+1\right)\left(y+1\right)^2}{3\sqrt[3]{x^2z^2}+1}+\dfrac{\left(y+1\right)\left(z+1\right)^2}{3\sqrt[3]{x^2y}+1}+\dfrac{\left(z+1\right)\left(x+1\right)^2}{3\sqrt[3]{y^2z^2}+1}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{\left(x+1\right)\left(y+1\right)^2}{3\sqrt[3]{x^2z^2}+1}=\dfrac{\left(x+1\right)\left(y+1\right)^2}{3\sqrt[3]{x\cdot z\cdot xz}+1}\ge\dfrac{\left(x+1\right)\left(y+1\right)^2}{x+z+xz+1}\)
\(=\dfrac{\left(x+1\right)\left(y+1\right)^2}{\left(x+1\right)\left(z+1\right)}=\dfrac{\left(y+1\right)^2}{z+1}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{\left(y+1\right)\left(z+1\right)^2}{3\sqrt[3]{x^2y^2}+1}\ge\dfrac{\left(z+1\right)^2}{x+1};\dfrac{\left(z+1\right)\left(x+1\right)^2}{3\sqrt[3]{y^2z^2}+1}\ge\dfrac{\left(x+1\right)^2}{y+1}\)
Cộng theo vế 3 BĐT trên rồi áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT\ge\dfrac{\left(x+y+z+3\right)^2}{x+y+z+3}=x+y+z+3=VP\)
