Ta chứng minh BĐT sau: \(x^3+y^3\ge xy\left(x+y\right)\) với \(x;y>0\)
Thật vậy: \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)
Đặt \(\left(x;y;z\right)=\left(a^3;b^3;c^3\right)\Rightarrow abc=1\)
\(P=\frac{1}{a^3+b^3+1}+\frac{1}{b^3+c^3+1}+\frac{1}{c^3+a^3+1}\le\frac{1}{ab\left(a+b\right)+1}+\frac{1}{bc\left(b+c\right)+1}+\frac{1}{ca\left(c+a\right)+1}\)
\(P\le\frac{abc}{ab\left(a+b\right)+abc}+\frac{abc}{bc\left(b+c\right)+abc}+\frac{abc}{ca\left(c+a\right)+abc}\)
\(P\le\frac{c}{a+b+c}+\frac{b}{a+b+c}+\frac{a}{a+b+c}=1\)
\(P_{max}=1\) khi \(a=b=c=1\) hay \(x=y=z=1\)