\(\dfrac{4}{\sqrt{x-2}}+\dfrac{1}{\sqrt{y-1}}+\dfrac{25}{\sqrt{z-5}}=16-\sqrt{x-2}-\sqrt{y-1}-\sqrt{z-5}\left(x>2;y>1;z>5\right)\)
\(\Leftrightarrow\left(\dfrac{4}{\sqrt{x-2}}+\sqrt{x-2}\right)+\left(\dfrac{1}{\sqrt{y-1}}+\sqrt{y-1}\right)+\left(\dfrac{25}{\sqrt{z-5}}+\sqrt{z-5}\right)=16\)
Ta có: \(\left\{{}\begin{matrix}x>2\Rightarrow\dfrac{4}{\sqrt{x-2}}>0;\sqrt{x-2}>0\\y>1\Rightarrow\dfrac{1}{\sqrt{y-1}}>0;\sqrt{y-1}>0\\z>5\Rightarrow\dfrac{25}{\sqrt{z-5}}>0;\sqrt{z-5}>0\end{matrix}\right.\)
Áp dụng bđt cô-si ta có:
\(VT\ge2\sqrt{\dfrac{4}{\sqrt{x-2}}\cdot\sqrt{x-2}}+2\sqrt{\dfrac{1}{\sqrt{y-1}}\cdot\sqrt{y-1}}+2\sqrt{\dfrac{25}{\sqrt{z-5}}+\sqrt{z-5}}\\ =2\sqrt{4}+2\sqrt{1}+2\sqrt{25}=2\cdot2+2\cdot1+2\cdot5=16\)
Dấu "=" xảy ra:
\(\left\{{}\begin{matrix}\dfrac{4}{\sqrt{x-2}}=\sqrt{x-2}\\\dfrac{1}{\sqrt{y-1}}=\sqrt{y-1}\\\dfrac{25}{\sqrt{z-5}}=\sqrt{z-5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=2\\z=30\end{matrix}\right.\left(tm\right)\)
