Question

[HUỲNH TÔ ÁI VÂN]

cho x, y, z duong. Chung minh rang 8\(\left(x^3+y^3+z^3\right)\ge\left(x+y\right)^3+\left(y+z\right)^3+\left(z+x\right)^3\)

Answer 1

[Lăng Y Vân]

Với x,y,z dương, áp dụng BĐT AM-GM:

\(\left\{{}\begin{matrix}x^3+x^3+y^3\ge3x^2y\\x^3+y^3+y^3\ge3xy^2\end{matrix}\right.\) \(\Rightarrow3\left(x^3+y^3\right)\ge3\left(x^2y+xy^2\right)\)

Tương tự:\(3\left(y^3+z^3\right)\ge3\left(y^2z+yz^2\right)\);\(3\left(x^3+z^3\right)\ge3\left(x^2z+xz^2\right)\)

Cộng vế theo vế:

\(\Leftrightarrow6\left(x^3+y^3+z^3\right)\ge3\left(x^2y+xy^2\right)+3\left(y^2z+yz^2\right)+3\left(x^2z+xz^2\right)\)

\(\Leftrightarrow8\left(x^3+y^3+z^3\right)\ge x^3+y^3+3xy\left(x+y\right)+y^3+z^3+3yz\left(y+z\right)+x^3+z^3+3xz\left(x+z\right)\) \(\Leftrightarrow8\left(x^3+y^3+z^3\right)\ge\left(x+y\right)^3+\left(y+z\right)^3+\left(x+z\right)^3\) (đpcm)