CMR: \(\left(x+z\right)\left(x+y\right)\ge2\) \(với\)\(x,y,z>0\)và \(\left(x+y+z\right)xyz=1\)
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26 tháng 7 2017
pt cái (x+y)(y+z)(z+x)=\(2xyz+z^2\left(x+y\right)+x^2\left(y+z\right)+y^2\left(x+z\right)\)
xét hiệu \(\left(x+y\right)\left(y+z\right)\left(x+z\right)-2\left(1+x+y+z\right)=2xyz+z^2\left(x+y\right)+y^2\left(x+z\right)+x^2\left(y+z\right)-2xyz-\left(x+y\right)-\left(y+z\right)-\left(x+y\right)\)\(z^2\left(x+y\right)\ge\left(x+y\right)\)(vì x;y;z>0)
tương tự
=> đpcm
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 3 2021
Ta có:
\(VT=2+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{y}{z}+\dfrac{x}{z}+\dfrac{z}{x}\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\ge\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
Ta có:
\(\dfrac{x}{y}+\dfrac{x}{y}+1\ge3\sqrt[3]{\dfrac{x^2}{y^2}}\)
Tương tự ...
Cộng lại ta có:
\(2\left(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\right)+6\ge3\left(\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\right)\)
\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\ge\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\)
Do đó ta chỉ cần chứng minh:
\(\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\ge\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
\(\Leftrightarrow\left(\sqrt[3]{\dfrac{x}{y}}-\sqrt[3]{\dfrac{x}{z}}\right)^2+\left(\sqrt[3]{\dfrac{y}{x}}-\sqrt[3]{\dfrac{y}{z}}\right)^2+\left(\sqrt[3]{\dfrac{z}{x}}-\sqrt[3]{\dfrac{z}{y}}\right)^2\ge0\) (luôn đúng)
AT
8 tháng 3 2019
\(VT=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(=2+\frac{z}{x}+\frac{y}{x}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}\)
Bài toán trở thành \(\frac{z}{x}+\frac{y}{x}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}\ge\frac{x+y+z}{3\sqrt{xyz}}\)
Áp dụng bất đẳng thức AM-GM:
\(\frac{z}{x}+\frac{z}{y}+\frac{z}{z}\ge3\sqrt[3]{\frac{z^3}{xyz}}=\frac{3z}{\sqrt[3]{xyz}}\)
Tương tự:
\(\frac{y}{x}+\frac{y}{z}+\frac{y}{y}\ge\frac{3y}{\sqrt[3]{xyz}}\)
\(\frac{x}{z}+\frac{x}{y}+\frac{x}{x}\ge\frac{3x}{\sqrt[3]{xyz}}\)
\(\Leftrightarrow VT+3\ge3+\frac{3}{\sqrt[3]{xyz}}\left(x+y+z\right)\)
\(\Leftrightarrow VT\ge\frac{3\left(x+y+z\right)}{\sqrt[3]{xyz}}\)\(\ge\frac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
Is it true?
14 tháng 8 2016
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.
25 tháng 2 2017
Áp dụng liên tiếp bđt AM-GM cho 2 số dương ta có:
A = \(\left(xyz+1\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\)\(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}=\left(xy+\frac{y}{x}\right)+\left(yz+\frac{z}{y}\right)+\)\(\left(xz+\frac{x}{z}\right)+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)\(\ge2\sqrt{xy.\frac{y}{x}}+2\sqrt{yz.\frac{z}{y}}+2\sqrt{xz.\frac{x}{z}}+\)\(+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(A\ge2y+2z+2x+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)\(=x+y+z+\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)+\left(z+\frac{1}{z}\right)\)
\(A\ge x+y+z+2\sqrt{x.\frac{1}{x}}+2\sqrt{y.\frac{1}{y}}+\)\(2\sqrt{z.\frac{1}{z}}=x+y+z+2.3=x+y+z+6\)(đpcm)
Dấu "=" xảy ra khi x = y = z = 1
23 tháng 6 2019
\(\frac{x}{1+x^2}=\frac{\frac{1}{x}}{\frac{1}{x^2}+1}=\frac{\frac{1}{x}}{\frac{1}{x^2}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}}=\frac{\frac{1}{x}}{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{z}+\frac{1}{x}\right)}\)
\(=\frac{xyz}{xy\left(\frac{1}{x}+\frac{1}{y}\right)zx\left(\frac{1}{z}+\frac{1}{x}\right)}=\frac{xyz}{\left(x+y\right)\left(z+x\right)}\)
Tương tự, ta cũng có: \(\frac{2y}{1+y^2}=\frac{2xyz}{\left(x+y\right)\left(y+z\right)}\)\(;\)\(\frac{3z}{1+z^2}=\frac{3xyz}{\left(y+z\right)\left(z+x\right)}\)
\(VT=\frac{xyz}{\left(x+y\right)\left(z+x\right)}+\frac{2xyz}{\left(x+y\right)\left(y+z\right)}+\frac{3xyz}{\left(y+z\right)\left(z+x\right)}\)
\(=\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\) ( đpcm )
ta có: \((x+y+z)xyz=1 \)
\((x+y+z)x={1\over yz} \)
hay\(x^2+xy+xz={1\over yz}\)
Ta có:\((x+y)(x+z)=x^2+xz+xy+yz={1\over yz}+yz\)
Áp dụng Bdt Cauchy là ra!!