Bài toán thiếu điều kiện \(x\ge1;y\ge2;z\ge3\)

Ta có : \(M=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)

Áp dụng bđt Cauchy, ta có : \(\frac{\sqrt{x-1}}{x}=\frac{\sqrt{\left(x-1\right).1}}{x}\le\frac{x-1+1}{2x}=\frac{x}{2x}=\frac{1}{2}\)

Tương tự : \(\frac{\sqrt{y-2}}{y}=\frac{\sqrt{\left(y-2\right).2}}{\sqrt{2}.y}\le\frac{y-2+2}{2\sqrt{2}.y}=\frac{y}{2\sqrt{2}y}=\frac{1}{2\sqrt{2}}\)

\(\frac{\sqrt{z-3}}{z}=\frac{\sqrt{\left(z-3\right).3}}{\sqrt{3}z}\le\frac{z-3+3}{2\sqrt{3}z}=\frac{z}{2\sqrt{3}z}=\frac{1}{2\sqrt{3}}\)

Cộng các bđt theo vế , được : \(M\le\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}\)

Dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}z-3=3\\y-2=2\\x-1=1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)

Vậy giá trị lớn nhất của M bằng \(\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}\) khi và chỉ khi (x;y;z) = (2;4;6)

\(=>A=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)

áp dụng BĐT AM-GM

\(=>\sqrt{x-1}\le\dfrac{x-1+1}{2}=\dfrac{x}{2}\)

\(=>\dfrac{\sqrt{x-1}}{x}\le\dfrac{\dfrac{x}{2}}{x}=\dfrac{1}{2}\left(1\right)\)

có \(\dfrac{\sqrt{y-2}}{y}=\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\)

\(=>\sqrt{\left(y-2\right)2}\le\dfrac{y-2+2}{2}=\dfrac{y}{2}\)

\(=>\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\le\dfrac{\dfrac{y}{2}}{\sqrt{2}.y}=\dfrac{1}{2\sqrt{2}}\left(2\right)\)

tương tự \(=>\dfrac{\sqrt{z-3}}{z}\le\dfrac{1}{2\sqrt{3}}\left(3\right)\)

(1)(2)(3)\(=>A\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)

Đề có vấn dề thì phải  căn thứ 2 ấy

Bài này CHTT  có thìphair

\(P=\frac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)

\(=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)

\(=\frac{2\sqrt{1.\left(x-1\right)}}{2x}+\frac{2\sqrt{2.\left(y-2\right)}}{2y\sqrt{2}}+\frac{2\sqrt{3.\left(z-3\right)}}{2z\sqrt{3}}\)

\(\le\frac{1+x-1}{2x}+\frac{2+y-2}{2y\sqrt{2}}+\frac{3+z-3}{2z\sqrt{3}}\)(cái này của BĐT cô-si thì phải)

\(=\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}1=x-1\\2=y-2\\3=z-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)

Vậy \(Min_{bt}=\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}\) khi \(\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)

Ta có: \(P=\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{zx}}{y+2\sqrt{zx}}+\frac{\sqrt{xy}}{z+2\sqrt{xy}}=\frac{1}{\frac{x}{\sqrt{yz}}+2}+\frac{1}{\frac{y}{\sqrt{zx}}+2}+\frac{1}{\frac{z}{\sqrt{xy}}+2}\)

Đặt \(\frac{x}{\sqrt{yz}}=c,\frac{y}{\sqrt{zx}}=t;\frac{z}{\sqrt{xy}}=k\left(c,t,k>0\right)\)thì ctk = 1

Ta cần tìm giá trị lớn nhất của \(P=\frac{1}{c+2}+\frac{1}{t+2}+\frac{1}{k+2}\)với ctk = 1

Dự đoán MaxP = 1 khi c = t = k = 1

Thật vậy: \(P=\frac{kt+2k+2t+4+ct+2c+2t+4+ck+2c+2k+4}{\left(c+2\right)\left(t+2\right)\left(k+2\right)}=\frac{\left(kt+tc+ck\right)+4\left(c+t+k\right)+12}{ctk+2\left(kt+tc+ck\right)+4\left(c+t+k\right)+8}\le\frac{\left(kt+tc+ck\right)+4\left(c+t+k\right)+12}{1+\left(kt+tc+ck\right)+3\sqrt[3]{\left(ctk\right)^2}+4\left(c+t+k\right)+8}=1\)Đẳng thức xảy ra khi x = y = z

Ta có: \(\frac{\sqrt{yz}}{x+2\sqrt{yz}}=\frac{1}{2}\left(1-\frac{x}{x+2\sqrt{yz}}\right)\le\frac{1}{2}\left(1-\frac{x}{x+y+z}\right)=\frac{1}{2}\left(\frac{y+z}{x+y+z}\right)\)(bđt cosi) (1)

CMTT: \(\frac{\sqrt{xz}}{y+2\sqrt{xz}}\le\frac{1}{2}\left(\frac{x+z}{x+y+z}\right)\)(2)

\(\frac{\sqrt{xy}}{z+2\sqrt{xy}}\le\frac{1}{2}\left(\frac{x+y}{x+y+z}\right)\)(3)

Từ (1), (2) và (3) cộng vế theo vế ta có:

\(\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{xz}}{y+2\sqrt{xz}}+\frac{\sqrt{xy}}{z+2\sqrt{xy}}\le\frac{1}{2}\left(\frac{y+z}{x+y+z}\right)+\frac{1}{2}\left(\frac{x+z}{x+y+z}\right)+\frac{1}{2}\left(\frac{x+y}{x+y+z}\right)\)

=> P \(\le\frac{1}{2}\left(\frac{y+z+x+z+x+y}{x+y+z}\right)=\frac{1}{2}\cdot\frac{2\left(x+y+z\right)}{x+y+z}=1\)

Dấu "=" xảy ra <=> x = y = z

Vậy MaxP = 1 <=> x = y = z

một bài khá hay :)

Ta có \(\frac{\sqrt{yz}}{x+2\sqrt{yz}}=1-\frac{x}{x+2\sqrt{yz}}\le1-\frac{x}{x+y+z}\left(1\right)\)

 Tương tự \(\frac{\sqrt{xz}}{y+2\sqrt{xz}}=1-\frac{y}{y+2\sqrt{xz}}\le1-\frac{y}{x+y+z}\left(2\right)\)

\(\frac{\sqrt{xy}}{z+2\sqrt{xy}}=1-\frac{z}{z+2\sqrt{xy}}\le1-\frac{z}{x+y+z}\left(3\right)\)

Cộng (1);(2);(3)

\(2P\le3-\frac{x+y+z}{x+y+z}=2\Rightarrow P\le1\)

Vậy \(minP=1\)Khi và chỉ khi \(x=y=z\)

\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)

\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)

\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)

\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)

\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)

\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)

Dấu = xảy ra khi \(x=y=z=9\)

 Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\) 

CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)  ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\) 

Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)

\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) 

Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\) 

Mặt khác :   \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)

Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)

" = " \(\Leftrightarrow x=y=z=9\)