Áp dụng BĐT Cauchy Schawrz dạng Engel, ta có:
\(\dfrac{1}{x}+\dfrac{16}{y}+\dfrac{9}{z}=\dfrac{1^2}{x}+\dfrac{4^2}{y}+\dfrac{3^2}{z}\)
\(\ge\dfrac{\left(1+4+3\right)^2}{x+y+z}=\dfrac{8^2}{x+y+z}\)(1)
\(\Leftrightarrow4\ge\dfrac{64}{x+y+z}\)
\(\Leftrightarrow4\left(x+y+z\right)\ge64\) (vì \(x+y+z>0\))
\(\Leftrightarrow x+y+z\ge16\)
Mà theo gt thì \(x+y+z\le16\)
\(\Rightarrow x+y+z=16\).
Dấu "=" xảy ra ở (1) khi và chỉ khi
\(\dfrac{1}{x}=\dfrac{4}{y}=\dfrac{3}{z}=\dfrac{1+4+3}{x+y+z}=\dfrac{8}{16}=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=8\\z=6\end{matrix}\right.\)(tm)
Vậy \(\left\{{}\begin{matrix}x=2\\y=8\\z=6\end{matrix}\right.\) thoả mãn đề bài.
