Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\frac{9}{1-2\left(xy+yz+xz\right)}+\frac{2}{xyz}=\frac{9}{\left(x+y+z\right)^2-2\left(xy+yz+xz\right)}+\frac{2\left(x+y+z\right)}{xyz}\)
\(=\frac{9}{x^2+y^2+z^2}+\frac{6\sqrt[3]{xyz}}{xyz}\ge\frac{9}{x^2+y^2+z^2}+\frac{18}{3\sqrt[3]{x^2y^2z^2}}\)
\(\ge\frac{9}{x^2+y^2+z^2}+\frac{36}{2\left(xy+yx+xz\right)}\ge9\left(\frac{1}{\left(x+y+z\right)^2}+\frac{2^2}{2\left(xy+yz=xz\right)}\right)\)
\(\ge\frac{81}{\left(x+y+z\right)^2=81}\)
Dấu = xảy ra khi x = y = z = 1/3
\(A=\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2x^2z^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)
\(A=\sqrt{\frac{x^2}{2xyz.yz+xz.xy}}+\sqrt{\frac{y^2}{2xyz.xz+xy.yz}}+\sqrt{\frac{z^2}{2xyz.xy+xz.yz}}\)
\(A=\sqrt{\frac{x^2}{yz\left(xy+yz+xz\right)+xz.xy}}+\sqrt{\frac{y^2}{xz\left(xy+yz+xz\right)+xy.yz}}+\sqrt{\frac{z^2}{xy\left(xy+yz+xz\right)+xz.yz}}\)
\(A=\sqrt{\frac{x^2}{\left(yz+xy\right)\left(yz+xz\right)}}+\sqrt{\frac{y^2}{\left(xz+xy\right)\left(xz+yz\right)}}+\sqrt{\frac{z^2}{\left(xy+yz\right)\left(xy+xz\right)}}\)
Áp dụng bđt \(\sqrt{ab}\le\frac{a+b}{2}\) ta có:
\(2A\le\frac{x}{yz+xy}+\frac{x}{yz+xz}+\frac{y}{xz+xy}+\frac{y}{xz+yz}+\frac{z}{xy+yz}+\frac{z}{xy+xz}\)
\(=\frac{x+z}{yz+xy}+\frac{x+y}{yz+xz}+\frac{y+z}{xz+xy}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Mà: \(xy+yz+xz=2xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(\Rightarrow2A\le2\Rightarrow A\le1."="\Leftrightarrow a=b=c=\frac{3}{2}\)
theo bài ra ta có: \(x^2+y^2+z^2=xyz\Rightarrow\frac{x^2+y^2+z^2}{xyz}=1.\)(vì x.y.z>o)
\(\Rightarrow\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}=1\)
mặt khác ta có: \(\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a+b\right)^2\ge4ab\) (với a>0;b>0)
\(\Leftrightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) (*)
Áp dụng bài toán (*) ta có: \(\frac{1}{x^2+yz}\le\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{yz}\right)\) \(\Rightarrow\frac{x}{x^2+yz}\le\frac{1}{4}\left(\frac{x}{x^2}+\frac{x}{yz}\right)\)
tương tự ta đc: \(\frac{y}{y^2+xz}\le\frac{1}{4}\left(\frac{y}{y^2}+\frac{y}{xz}\right)\) ; \(\frac{z}{z^2+xy}\le\frac{1}{4}\left(\frac{z}{z^2}+\frac{z}{xy}\right)\)
\(\Rightarrow A\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}\right)=\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1\right)\) (vì \(\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}=1\))
mà \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+xz}{xyz}\le\frac{x^2+y^2+z^2}{xyz}=1\)
\(\Rightarrow A\le\frac{1}{4}\left(1+1\right)=\frac{1}{2}\)
Vậy GTLN của A là 1/2 khi x=y=z