\(\left\{\begin{matrix}\sqrt{2x-3}=a\\\sqrt{y-2}=b\\\sqrt{3z-1}=c\end{matrix}\right.\) \(\left\{\begin{matrix}a>0\\b>0\\c>0\end{matrix}\right.\)
\(Q=\left(\frac{1}{a}+a\right)+\left(\frac{4}{b}+b\right)+\left(\frac{1}{c}+c\right)\)
\(\left\{\begin{matrix}\frac{1}{a}+a\ge2\forall a>0\\\frac{4}{b}+b\ge4\forall b>0\\\frac{16}{c}+c\ge8\forall c>0\end{matrix}\right.\) đẳng thức khi \(\left\{\begin{matrix}a=1\\b=2\\c=4\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=2\\y=6\\z=\frac{17}{3}\end{matrix}\right.\)
cộng lại \(Q\ge14\)
Do Z không nguyên ta phải xét
f(z)=\(\frac{1}{\sqrt{3z-1}}+\sqrt{3z-1}\) \(f\left(6\right)=\frac{16}{\sqrt{3.6-1}}+\sqrt{3.6-1}=\frac{16+17}{\sqrt{17}}=\frac{33}{\sqrt{17}}\)
\(f\left(5\right)=\frac{16}{\sqrt{3.5-1}}+\sqrt{3.5-1}=\frac{16+14}{\sqrt{14}}=\frac{30}{\sqrt{14}}\)
\(\left[f\left(6\right)\right]^2-\left[f\left(5\right)\right]^2=\frac{14.33^2.-17.30^2}{17.14}=\frac{\left(17-3\right).33^2-17.\left(33-3\right)^2}{17.14}=\frac{17.33^2-3.33^2-\left[17.33^2-6.33.17+17.9\right]}{17.14}=\frac{-3.33^2-17.9+6.33.17}{17.14}=\frac{6.33\left(17-3.33\right)-17.9}{17.14}< 0\)
\(\Rightarrow f\left(6\right)< f\left(5\right)\)
\(Q_{min\left(x,y,z\in Z\right)}=2+6+\frac{33}{\sqrt{17}}=8+\frac{33\sqrt{17}}{17}\)
