Question

Tìm giá trị nhỏ nhất:

\(\dfrac{5}{\left|x+y\right|^2+\left(y+z\right)^3+\left|z+x\right|^4}\)

Answer 1

\(\dfrac{5}{\left|x-y\right|^2+\left(y-z\right)^2+\left|z-x\right|^4+6}\)

Cái này cũng làm tương tự như cái kia thôi:
Ta có:

\(\left\{{}\begin{matrix}\left|x-y\right|^2\ge0\\\left(y-z\right)^2\ge0\\\left|z-x\right|^4\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|x-y\right|^2+\left(y-z\right)^2+\left|z-x\right|^4\ge0\)

\(\Leftrightarrow\left|x-y\right|^2+\left(y-z\right)^2+\left|z-x\right|^4+6\ge6\)

\(A=\dfrac{5}{\left|x-y\right|^2+\left(y-z\right)^2+\left|z-x\right|^4+6}\ge\dfrac{5}{6}\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-y\right|^2=0\\\left(y-z\right)^2=0\\\left|z-x\right|^4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\Leftrightarrow x=y=z\)

Vậy

Answer 2

\(\dfrac{5}{\left|x-y\right|^2+\left(y-z\right)^2+\left|z-x\right|^4+6}\)