ĐK : \(x\ge3;y\ge1;z\ge665\)
\(\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\sqrt{x-3}-\sqrt{y-1}-\sqrt{z-665}\)
\(\Leftrightarrow\left(\dfrac{16}{\sqrt{x-3}}+\sqrt{x-3}\right)+\left(\dfrac{4}{\sqrt{y-1}}+\sqrt{y-1}\right)+\left(\dfrac{1225}{\sqrt{z-665}}+\sqrt{z-665}\right)=82\)
Theo BĐT Cô Si cho các số dương ta có :
\(\left\{{}\begin{matrix}\dfrac{16}{\sqrt{x-3}}+\sqrt{x-3}\ge2\sqrt{\dfrac{16\sqrt{x-3}}{\sqrt{x-3}}}=2\sqrt{16}=8\\\dfrac{4}{\sqrt{y-1}}+\sqrt{y-1}\ge2\sqrt{\dfrac{4\sqrt{y-1}}{\sqrt{y-1}}}=2\sqrt{4}=4\\\dfrac{1225}{\sqrt{z-665}}+\sqrt{z-665}\ge2\sqrt{\dfrac{1225\sqrt{z-665}}{\sqrt{z-665}}}=2\sqrt{1225}=70\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{16}{\sqrt{x-3}}+\sqrt{x-3}\right)+\left(\dfrac{4}{\sqrt{y-1}}+\sqrt{y-1}\right)+\left(\dfrac{1225}{\sqrt{z-665}}+\sqrt{z-665}\right)\ge82\)
Dấu \("="\) hiển nhiên xảy ra khi :
\(\left\{{}\begin{matrix}\dfrac{16}{\sqrt{x-3}}=\sqrt{x-3}\\\dfrac{4}{\sqrt{y-1}}=\sqrt{y-1}\\\dfrac{1225}{\sqrt{z-665}}=\sqrt{z-665}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=16\\y-1=4\\z-665=1225\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=19\\y=5\\z=1890\end{matrix}\right.\)
