\(1\) \(cách\) \(làm\) \(khác\) \(B=\dfrac{1}{\dfrac{x}{\sqrt{yz}}+2}+\dfrac{1}{\dfrac{y}{\sqrt{xz}}+2}+\dfrac{1}{\dfrac{z}{\sqrt{xy}}+2}\)
\(đặt:\left(\dfrac{x}{\sqrt{yz}};\dfrac{y}{\sqrt{xz}};\dfrac{z}{\sqrt{xy}}\right)=\left(a;b;c\right)\Rightarrow abc=1\)
\(\Rightarrow B=\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\)
\(ta\) \(đi\) \(cm:B\le1\Leftrightarrow\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\le1\)
\(\Leftrightarrow\left(b+2\right)\left(c+2\right)+\left(a+2\right)\left(c+2\right)+\left(a+2\right)\left(b+2\right)\le\left(a+2\right)\left(b+2\right)\left(c+2\right)\)
\(\Leftrightarrow ab+bc+ca+abc-4\ge0\Leftrightarrow ab+bc+ca\ge3\)
điều này đúng theo cosi \(ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}=3\)
\(dấu"="\Leftrightarrow a=b=c=1\Leftrightarrow x=y=z\)
\(x,y,z>0\)
\(B=\dfrac{\sqrt{yz}}{x+2\sqrt{yz}}+\dfrac{\sqrt{zx}}{y+2\sqrt{zx}}+\dfrac{\sqrt{xy}}{z+2\sqrt{xy}}\)
\(\Rightarrow2B=\dfrac{2\sqrt{yz}}{x+2\sqrt{yz}}+\dfrac{2\sqrt{zx}}{y+2\sqrt{zx}}+\dfrac{2\sqrt{xy}}{z+2\sqrt{xy}}\)
\(=1+1+1-\left(\dfrac{x}{x+2\sqrt{yz}}+\dfrac{y}{y+2\sqrt{zx}}+\dfrac{z}{z+2\sqrt{xy}}\right)\)
\(=3-\left(\dfrac{x^2}{x^2+2x\sqrt{yz}}+\dfrac{y^2}{y^2+2y\sqrt{zx}}+\dfrac{z^2}{z^2+2z\sqrt{xy}}\right)\)
\(\le3-\dfrac{\left(x+y+z\right)^2}{x^2+2x\sqrt{yz}+y^2+2y\sqrt{zx}+z^2+2z\sqrt{xy}}\)
\(\le3-\dfrac{\left(x+y+z\right)^2}{x^2+2x.\dfrac{y+z}{2}+y^2+2y.\dfrac{z+x}{2}+z^2+2z.\dfrac{x+y}{2}}\)
\(=3-\dfrac{\left(x+y+z\right)^2}{x^2+xy+xz+y^2+yz+yx+z^2+zx+zy}\)
\(=3-\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3\)
\(\Rightarrow B\le\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z>0\)
Vậy \(MaxB=\dfrac{3}{2}\)
