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ĐKXĐ : \(\hept{\begin{cases}x\ge0\\y\ge0\\x\ne y\end{cases}}\)
a) \(C=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{x}-y\sqrt{y}}{y-x}\right)\)
\(C=\frac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\frac{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)-x\sqrt{x}+y\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(C=\frac{x+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{x\sqrt{x}+x\sqrt{y}-y\sqrt{x}-y\sqrt{y}-x\sqrt{x}+y\sqrt{y}}\)
\(C=\frac{\left(x+y-\sqrt{xy}\right)\left(\sqrt{x}-\sqrt{y}\right)}{x\sqrt{y}-y\sqrt{x}}\)
\(C=\frac{\left(x+y-\sqrt{xy}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}\)
\(C=\frac{x+y-\sqrt{xy}}{\sqrt{xy}}\)
b)Giả sử \(C>1\)
\(\Leftrightarrow\frac{x+y-\sqrt{xy}}{\sqrt{xy}}>1\)
\(\Leftrightarrow\frac{x+y-\sqrt{xy}-\sqrt{xy}}{\sqrt{xy}}>0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{xy}}>0\)( luôn đúng với mọi \(\hept{\begin{cases}x\ge0\\y\ge0\\x\ne y\end{cases}}\))
Nhầm ĐKXĐ :\(\hept{\begin{cases}x>0\\y>0\\x\ne y\end{cases}}\)
a) Ta có: \(C=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{x}-y\sqrt{y}}{y-x}\right)\)
\(=\frac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)
\(=\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}:\left(\frac{x\sqrt{x}+x\sqrt{y}-y\sqrt{x}-y\sqrt{y}-\left(x\sqrt{x}-y\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)
\(=\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}:\frac{x\sqrt{x}+x\sqrt{y}-y\sqrt{x}-y\sqrt{y}-x\sqrt{x}+y\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\cdot\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{x\sqrt{y}-y\sqrt{x}}\)
\(=\frac{\left(x-\sqrt{xy}+y\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}\)
\(=\frac{x-\sqrt{xy}+y}{\sqrt{xy}}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\y>0\\x\ne y\end{matrix}\right.\)
Ta có: \(C-1=\frac{x-\sqrt{xy}+y}{\sqrt{xy}}-1\)
\(=\frac{x-\sqrt{xy}+y-\sqrt{xy}}{\sqrt{xy}}\)
\(=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{xy}}>0\forall x,y\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow C-1>0\)
hay C>1(đpcm)
Áp dụng BĐT Cô si ta có:
\(x+y\ge2\sqrt{xy}=2\cdot\frac{1}{\sqrt{z}};y+z\ge2\sqrt{yz}=2\cdot\frac{1}{\sqrt{x}};z+x\ge2\sqrt{xz}=2\cdot\frac{1}{\sqrt{y}}.\)( vì xyz=1)
=> P\(\ge\)\(\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}\)+ \(\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)
Đặt \(\hept{\begin{cases}a=y\sqrt{y}+2z\sqrt{z}\\b=z\sqrt{z}+2x\sqrt{x}\\c=x\sqrt{x}+2y\sqrt{y}\end{cases}\left(a;b;c\ge0\right)}\)<=> \(\hept{\begin{cases}4a+b=2c+9z\sqrt{z}\\4b+c=2a+9x\sqrt{x}\\4c+a=2b+9y\sqrt{y}\end{cases}}\)
<=> \(\hept{\begin{cases}z\sqrt{z}=\frac{4a+b-2c}{9}\\x\sqrt{x}=\frac{4b+c-2a}{9}\\y\sqrt{y}=\frac{4c+a-2b}{9}\end{cases}}\)
Do đó:
P \(\ge\)\(\frac{2}{9}\cdot\left(\frac{4a+b-2c}{c}+\frac{4b+c-2a}{a}+\frac{4c+a-2b}{b}\right)\)
<=> P \(\ge\)\(\frac{2}{9}\left(4\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)+\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}\right)-6\right)\)
<=> P \(\ge\frac{2}{9}\cdot\left(4\cdot3\cdot\sqrt[3]{\frac{a}{c}\cdot\frac{b}{a}\cdot\frac{c}{b}}+3\cdot\sqrt[3]{\frac{b}{c}\cdot\frac{c}{a}\cdot\frac{a}{b}}-6\right)\)( Áp dụng BĐT Cô si cho 3 số ko âm)
<=> P \(\ge\frac{2}{9}\left(12+3-6\right)=2\)( đpcm)
Dấu = khi x=y=z=1.
\(\frac{\left(x+y\right)^2}{2}+\frac{x+y}{4}\ge x\sqrt{y}+y\sqrt{x}\)
\(\Leftrightarrow2\left(x+y\right)^2+x+y-4x\sqrt{y}-4y\sqrt{x}\ge0\)
\(\Leftrightarrow2x^2+4xy+2y^2+x+y-4x\sqrt{y}-4y\sqrt{x}\ge0\)
\(\Leftrightarrow\left(4xy-4x\sqrt{y}+x\right)+\left(4xy-4y\sqrt{x}+y\right)+\left(2x^2-4xy+2y^2\right)\ge0\)
\(\Leftrightarrow\left(2\sqrt{xy}-\sqrt{x}\right)^2+\left(2\sqrt{xy}-\sqrt{y}\right)^2+2\left(x-y\right)^2\ge0\)
\(\Rightarrow\)ĐPCM
Đề có vẻ thiếu một giả thiết nào đó.