Question

Tìm GTLN:

\(\dfrac{1}{\left(x-y\right)^2+\left(y-z\right)^4+\left(z-x\right)^6+5}\)

Answer 1

\(\dfrac{1}{\left(x-y\right)^2+\left(y-z\right)^4+\left(z-x\right)^6+5}\)

\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-z\right)^4\ge0\\\left(z-x\right)^6\ge0\end{matrix}\right.\)

\(\left(x-y\right)^2+\left(y-z\right)^4+\left(z-x\right)^6\ge0\)

\(\left(x-y\right)^2+\left(y-z\right)^4+\left(z-x\right)^6+5\ge5\)

\(\dfrac{1}{\left(x-y\right)^2+\left(y-z\right)^4+\left(z-x\right)^6+5}\ge\dfrac{1}{5}\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\) \(\Rightarrow x=y=z\)