Đặt \(a=\sqrt{2x-3}\) ; \(b=\sqrt{y-2}\) ; \(c=\sqrt{3z-1}\) (\(a,b,c>0\))

Ta có : \(\frac{1}{a}+\frac{4}{b}+\frac{16}{c}+a+b+c=14\)

\(\Leftrightarrow\left(\sqrt{2x-3}+\frac{1}{\sqrt{2x-3}}-2\right)+\left(\sqrt{y-2}+\frac{4}{\sqrt{y-2}}-4\right)+\left(\sqrt{3z-1}+\frac{16}{\sqrt{3z-1}}-8\right)=0\)

\(\Leftrightarrow\left[\frac{\left(2x-3\right)-2\sqrt{2x-3}+1}{\sqrt{2x-3}}\right]+\left[\frac{\left(y-2\right)-4\sqrt{y-2}+4}{\sqrt{y-2}}\right]+\left[\frac{\left(3z-1\right)-8\sqrt{3z-1}+16}{\sqrt{3z-1}}\right]=0\)

\(\Leftrightarrow\frac{\left(\sqrt{2x-3}-1\right)^2}{\sqrt{2x-3}}+\frac{\left(\sqrt{y-2}-2\right)^2}{\sqrt{y-2}}+\frac{\left(\sqrt{3z-1}-4\right)^2}{\sqrt{3z-1}}=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{2x-3}-1\right)^2=0\\\left(\sqrt{y-2}-2\right)^2=0\\\left(\sqrt{3z-1}-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=\frac{17}{3}\end{cases}}}\)(TMĐK)

Vậy : \(\left(x;y;z\right)=\left(2;6;\frac{17}{3}\right)\)

Phần đặt ẩn a,b,c bạn bỏ đi nhé ^^

Áp dụng bđt AM-GM cho 2 số dương ta có:

\(\left(\frac{1}{\sqrt{2x-3}}+\sqrt{2x-3}\right)+\left(\frac{4}{\sqrt{y-2}}+\sqrt{y-2}\right)\)\(+\left(\frac{16}{\sqrt{3z-1}}+\sqrt{3z-1}\right)\ge\)\(2\sqrt{\frac{1}{\sqrt{2x-3}}.\sqrt{2x-3}}+2\sqrt{\frac{4}{\sqrt{y-2}}.\sqrt{y-2}}\)\(+2\sqrt{\frac{16}{\sqrt{3z-1}}.\sqrt{3z-1}}=2.1+2.2+2.4=14\)

Dau "=" xay ra khi \(\left\{\begin{matrix}\frac{1}{\sqrt{2x-3}}=\sqrt{2x-3}\\\frac{4}{\sqrt{y-2}}=\sqrt{y-2}\\\frac{16}{\sqrt{3z-1}}=\sqrt{3z-1}\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}2x-3=1\\y-2=4\\3z-1=16\end{matrix}\right.\)=> \(\left\{\begin{matrix}x=1\\y=6\\z=\frac{17}{3}\end{matrix}\right.\) (không TM z nguyên dương)

Vay ...

Đật 3 cái mẫu bên VT lần lượt là x,y,z rồi áp dụng C-S dạng engel

Để dễ nhìn ta đặt \(\hept{\begin{cases}\sqrt{2x-3}=a\\\sqrt{y-2}=b\\\sqrt{3z-1}=c\end{cases}\left(a,b,c\ge0\right)}\)

Vậy BĐT đầu tương đương \(T=\frac{1}{a}+\frac{4}{b}+\frac{16}{c}+a+b+c\)

Áp dụng BĐT C-S dạng Engel ta có:

\(\frac{1}{a}+\frac{4}{b}+\frac{16}{c}=\frac{1^2}{a}+\frac{2^2}{b}+\frac{4^2}{c}\ge\frac{\left(1+2+4\right)^2}{a+b+c}=\frac{49}{a+b+c}\)

Tiếp tục dùng AM-GM ta có: \(VT\ge\frac{49}{a+b+c}+\left(a+b+c\right)\ge2\sqrt{\frac{49}{a+b+c}\cdot\left(a+b+c\right)}=2\sqrt{49}=14\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}a=1\\b=2\\c=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=\frac{17}{3}\end{cases}}\)

nhìn qua thì chắc AM-GM+Cauchy-schwarz chắc thế :)

\(\left\{\begin{matrix}\sqrt{2x-3}=a\\\sqrt{y-2}=b\\\sqrt{3z-1}=c\end{matrix}\right.\) \(\left\{\begin{matrix}a>0\\b>0\\c>0\end{matrix}\right.\)

\(Q=\left(\frac{1}{a}+a\right)+\left(\frac{4}{b}+b\right)+\left(\frac{1}{c}+c\right)\)

\(\left\{\begin{matrix}\frac{1}{a}+a\ge2\forall a>0\\\frac{4}{b}+b\ge4\forall b>0\\\frac{16}{c}+c\ge8\forall c>0\end{matrix}\right.\) 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)=\(\frac{1}{\sqrt{3z-1}}+\sqrt{3z-1}\) \(f\left(6\right)=\frac{16}{\sqrt{3.6-1}}+\sqrt{3.6-1}=\frac{16+17}{\sqrt{17}}=\frac{33}{\sqrt{17}}\)

\(f\left(5\right)=\frac{16}{\sqrt{3.5-1}}+\sqrt{3.5-1}=\frac{16+14}{\sqrt{14}}=\frac{30}{\sqrt{14}}\)

\(\left[f\left(6\right)\right]^2-\left[f\left(5\right)\right]^2=\frac{14.33^2.-17.30^2}{17.14}=\frac{\left(17-3\right).33^2-17.\left(33-3\right)^2}{17.14}=\frac{17.33^2-3.33^2-\left[17.33^2-6.33.17+17.9\right]}{17.14}=\frac{-3.33^2-17.9+6.33.17}{17.14}=\frac{6.33\left(17-3.33\right)-17.9}{17.14}< 0\)

\(\Rightarrow f\left(6\right)< f\left(5\right)\)

\(Q_{min\left(x,y,z\in Z\right)}=2+6+\frac{33}{\sqrt{17}}=8+\frac{33\sqrt{17}}{17}\)