\(2x+yz=x\left(x+y+z\right)+yz=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow\sqrt{2x+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{x+y+x+z}{2}=\dfrac{2x+y+z}{2}\) ( BĐT Cauchy )
Tương tự : \(\sqrt{2y+xz}\le\dfrac{2y+x+z}{2};\sqrt{2z+xy}\le\dfrac{2z+x+y}{2}\)
Suy ra : \(P\le\dfrac{1}{2}\left(4x+4y+4z\right)=2\left(x+y+z\right)=4\) ( do x + y + z = 2 )
" = " \(\Leftrightarrow x=y=z=\dfrac{2}{3}\)
Ta có:
\(\sqrt{2x+yz}\)
\(=\sqrt{\left(x+y+z\right)x+yz}\)
\(=\sqrt{\left(x^2+xy\right)+\left(xz+yz\right)}\)
\(=\sqrt{x\left(x+y\right)+z\left(x+y\right)}\)
\(=\sqrt{\left(x+y\right)\left(x+z\right)}\)
Áp dụng bất đẳng thức Cô-si, ta có:
\(\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{\left(x+y\right)+\left(x+z\right)}{2}=\dfrac{2x+y+z}{2}\Leftrightarrow\sqrt{2x+yz}\le\dfrac{2x+y+z}{2}\)
Tương tự, ta có: \(\sqrt{2y+xz}\le\dfrac{2y+x+z}{2}\) và \(\sqrt{2z+xy}\le\dfrac{2z+x+y}{2}\)
Từ đó: \(P\le\dfrac{2x+y+z}{2}+\dfrac{2y+x+z}{2}+\dfrac{2z+x+y}{2}\)
\(\Leftrightarrow P\le\dfrac{2x+y+z+2y+x+z+2z+x+y}{2}\)
\(\Leftrightarrow P\le\dfrac{4x+4y+4z}{2}\)
\(\Leftrightarrow P\le4\)
Đẳng thức xảy ra \(\Leftrightarrow x=y=z=\dfrac{2}{3}.\)
Vậy \(MaxP=4\Leftrightarrow x=y=z=\dfrac{2}{3}.\)
\(P=\sqrt{2x+yz}+\sqrt{2y+zx}+\sqrt{2z+xy}\)
\(=\sqrt{\dfrac{9}{16}}\left(\sqrt{\left(2x+yz\right).\dfrac{16}{9}}+\sqrt{\left(2y+zx\right).\dfrac{16}{9}}+\sqrt{\left(2z+xy\right).\dfrac{16}{9}}\right)\)
\(\le\dfrac{3}{4}\left[\dfrac{\left(2x+yz\right)+\dfrac{16}{9}}{2}+\dfrac{\left(2y+zx\right)+\dfrac{16}{9}}{2}+\dfrac{\left(2z+xy\right)+\dfrac{16}{9}}{2}\right]\)
\(=\dfrac{3}{4}\left[x+y+z+\dfrac{xy+yz+zx}{2}+\dfrac{8}{3}\right]\)
\(=\dfrac{3}{4}\left(2+\dfrac{xy+yz+zx}{2}+\dfrac{8}{3}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{14}{3}+\dfrac{xy+yz+zx}{2}\right)\)
\(\le\dfrac{3}{4}\left[\dfrac{14}{3}+\dfrac{\dfrac{\left(x+y+z\right)^2}{3}}{2}\right]\)
\(=\dfrac{3}{4}\left(\dfrac{14}{3}+\dfrac{2^2}{6}\right)=4\)
- Dấu "=" xảy ra khi \(x=y=z=\dfrac{2}{3}\)
