Đặt \(\left(x^3;y^3;z^3\right)=\left(a;b;c\right)\left(x,y,z>0\right)\)
\(\Rightarrow xyz=1\)
Ta cần chứng minh
\(\dfrac{1}{x^3+y^3+1}+\dfrac{1}{y^3+z^3+1}+\dfrac{1}{z^3+x^3+1}\le1\)
Áp dụng AM-GM, ta có: \(x^3+y^3+1=\left(x+y\right)\left(x^2-xy+y^2\right)+xyz\)
\(\ge\left(x+y\right)xy+xyz=xy\left(x+y+z\right)\)
\(\Rightarrow\dfrac{1}{x^3+y^3+1}\le\dfrac{1}{xy\left(x+y+z\right)}\)
Tương tự: \(\dfrac{1}{y^3+z^3+1}\le\dfrac{1}{yz\left(x+y+z\right)}\)
\(\dfrac{1}{z^3+x^3+1}\le\dfrac{1}{zx\left(x+y+z\right)}\)
Cộng vế theo vế, ta được
\(....\le\dfrac{1}{x+y+z}\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=\dfrac{1}{x+y+z}.\dfrac{x+y+z}{xyz}=\dfrac{1}{xyz}=1\)
Vậy ta có đpcm
Đẳng thức xảy ra khi a=b=c=1