x+y=z
y+z=x
x+z=y
xyz=1
tìm x y z
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AT
16 tháng 7 2021
2. \(P=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}\) (BĐT Cauchy-Schwarz)
\(=\dfrac{1}{2}\)
\(\Rightarrow P_{min}=\dfrac{1}{2}\) khi \(\dfrac{x}{y+z}=\dfrac{y}{z+x}=\dfrac{z}{x+y}\Rightarrow x=y=z=\dfrac{1}{3}\)
16 tháng 7 2021
1, đặt \(x^2+x=t\)
=>\(A=t\left(t-4\right)=t^2-4t=t^2-4t+4-4\)
\(=>A=\left(t-2\right)^2-4\ge-4\) dấu"=' xảy ra\(t=2\)
\(=>x^2+x=2< =>x^2+x-2=0\)
\(< =>x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{9}{4}=0\)
\(< =>\left(x+\dfrac{1}{2}\right)^2-\left(\dfrac{3}{2}\right)^2=0< =>\left(x-1\right)\left(x+2\right)=0\)
\(=>\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\) vậy Amin=-4<=>\(\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
B2
\(=>P=\dfrac{x^2}{y+z}+\dfrac{y+z}{4}+\dfrac{y^2}{x+z}+\dfrac{x+z}{4}+\dfrac{z^2}{x+y}+\dfrac{x+y}{4}\)
\(-\left(\dfrac{y+z+x+z+x+y}{4}\right)\)
áp dụng BDT AM-GM
\(=>\dfrac{x^2}{y+z}+\dfrac{y+z}{4}\ge2\sqrt{\dfrac{x^2}{4}}=x^{ }\left(1\right)\)
\(\)tương tự \(=>\dfrac{y^2}{x+z}+\dfrac{x+z}{4}\ge y\left(2\right)\)
\(=>\dfrac{z^2}{x+y}+\dfrac{x+y}{4}\ge z\left(3\right)\)
(1)(2)(3) \(=>P\ge x+y+z-\dfrac{1}{2}.x+y+z=1-\dfrac{1}{2}=\dfrac{1}{2}\)
dấu"=" xảy ra<=>x=y=z=1/3
T
8 tháng 8 2021
Sử dụng bất đẳng thức Minkovski, ta có:
\(P = \sqrt {{{\left( {x + y + z} \right)}^2} + {{\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)}^2}} \)
\( \ge \sqrt {\left[ {{{\left( {x + y + z} \right)}^2} + \frac{1}{{{{\left( {x + y + z} \right)}^2}}}} \right] + \frac{{80}}{{{{\left( {x + y + z} \right)}^2}}}} \)
\(\ge \sqrt{2+\dfrac{80}{1}} =\sqrt{82}\)
Đẳng thức xảy ra khi \(x=y=z=\dfrac{1}{3}.\)
Kết luận ...
8 tháng 8 2021
\(\sqrt{x^2+\dfrac{1}{x^2}}=\dfrac{1}{\sqrt{82}}\sqrt{\left(1^2+9^2\right)\left(x^2+\dfrac{1}{x^2}\right)}\ge\dfrac{1}{\sqrt{82}}\left(x+\dfrac{9}{x}\right)\)
tương tự với \(\sqrt{y^2+\dfrac{1}{y^2}};\sqrt{z^2+\dfrac{1}{z^2}}\)
\(=>P\ge\dfrac{1}{\sqrt{81}}\left(x+\dfrac{9}{x}+y+\dfrac{9}{y}+z+\dfrac{9}{z}\right)\)
có \(x+\dfrac{9}{x}=x+\dfrac{1}{9x}+\dfrac{80}{9x}\ge2\sqrt{\dfrac{1}{9}}+\dfrac{80}{9x}\)
tương tự với \(y+\dfrac{9}{y};z+\dfrac{9}{z}\)
\(=>P\ge\dfrac{1}{\sqrt{82}}\left[2\sqrt{\dfrac{1}{9}}.3+\dfrac{\left(\sqrt{80}+\sqrt{80}+\sqrt{80}\right)^2}{9\left(x+y+z\right)}\right]=\dfrac{1}{\sqrt{82}}.82=\sqrt{82}\)
dấu"=" xảy ra<=>x=y=z=1/3
27 tháng 1 2022
Ta có: \(\sqrt{\left(x^2+\dfrac{1}{y^2}\right)\left(1+81\right)}\ge\sqrt{\left(x+\dfrac{9}{y}\right)^2}\)
=> \(\sqrt{x^2+\dfrac{1}{y^2}}\ge\dfrac{x+\dfrac{9}{y}}{\sqrt{82}}\)
Tương tự => \(\left\{{}\begin{matrix}\sqrt{y^2+\dfrac{1}{z^2}}\ge\dfrac{y+\dfrac{9}{z}}{\sqrt{82}}\\\sqrt{z^2+\dfrac{1}{x^2}}\ge\dfrac{z+\dfrac{9}{x}}{\sqrt{82}}\end{matrix}\right.\)
=> \(P\ge\dfrac{\left(x+y+z\right)+9\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}{\sqrt{82}}\)
Mà x + y + z = 1
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}=9\)
=> \(P\ge\sqrt{82}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{3}\)
HQ
1
5 tháng 8 2020
a/ \(x\left(y-z\right)+y\left(z-x\right)+z\left(x-y\right)\)
\(=xy-xz+yz-xy+zx-yz\)
\(=0\)
Vậy...
b/ \(x\left(y+z-yz\right)-y\left(z+x-zx\right)+zy+x\)
\(=xy+xz-xyz-yz-xy+xyz+zy+x\)
\(=x\)
Vậy....
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 11 2021
\(\left\{{}\begin{matrix}x;y;z\ge0\\x+y+z=1\end{matrix}\right.\) \(\Rightarrow0\le x;y;z\le1\)
\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\\z^2\le z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x^2+x+1\le x^2+2x+1\\2y^2+y+1\le y^2+2y+1\\2z^2+z+1\le z^2+2z+1\end{matrix}\right.\)
\(\Rightarrow P\le\sqrt{\left(x+1\right)^2}+\sqrt{\left(y+1\right)^2}+\sqrt{\left(z+1\right)^2}=x+y+z+3=4\)
\(P_{max}=4\) khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 1 2023
\(D\le\dfrac{1}{2}\left(1+\dfrac{x}{1+yz}\right)+\dfrac{1}{2}\left(1+\dfrac{y}{1+zx}\right)+\dfrac{z}{2+2xy}\)
\(=1+\dfrac{x}{2\left(1+yz\right)}+\dfrac{y}{2\left(1+zx\right)}+\dfrac{z}{2\left(1+xy\right)}\)
Do \(0\le x;y;z\le1\)
\(\Rightarrow\left(1-x\right)\left(1-y\right)\ge0\Leftrightarrow xy+1\ge x+y\)
\(\Leftrightarrow2\left(xy+1\right)\ge xy+1+x+y\ge x+y+z\)
\(\Rightarrow\dfrac{z}{2\left(1+xy\right)}\le\dfrac{z}{x+y+z}\)
Tương tự: \(\dfrac{x}{2\left(1+yz\right)}\le\dfrac{x}{x+y+z}\) ; \(\dfrac{y}{2\left(1+zx\right)}\le\dfrac{y}{x+y+z}\)
Cộng vế:
\(P\le1+\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}=2\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;0\right)\)
NV
1
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 11 2021
\(A=2\left(x^2+y^2\right)+\left(8y^2+\dfrac{1}{2}z^2\right)+\left(8x^2+\dfrac{1}{2}z^2\right)\ge2.2\sqrt{x^2y^2}+2\sqrt{8x^2.\dfrac{1}{2}z^2}+2.\sqrt{8x^2.\dfrac{1}{2}z^2}=4\left(xy+yz+zx\right)=4\)
\(A_{min}=4\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{3};\dfrac{1}{3};\dfrac{4}{3}\right)\)
x=y=z=rỗng