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Vì x+y+z=1
=>\(\frac{1}{x}=1+\frac{y}{x}+\frac{z}{x}\)
\(\frac{1}{y}=1+\frac{x}{y}+\frac{z}{y}\)
\(\frac{1}{z}=1+\frac{x}{z}+\frac{y}{z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\left(1+\frac{y}{x}+\frac{z}{x}\right)+\left(1+\frac{x}{y}+\frac{z}{y}\right)+\left(1+\frac{x}{z}+\frac{y}{z}\right)\)
\(=\left(1+1+1\right)+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\)
\(=3+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\)
Ta có: a,b,c là các số dương nên theo bất đẳng thức Cô-Si:
\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\)
\(\frac{y}{z}+\frac{z}{y}\ge2\sqrt{\frac{y}{z}.\frac{z}{y}}=2\)
\(\frac{x}{z}+\frac{z}{x}\ge2\sqrt{\frac{x}{z}.\frac{z}{x}}=2\)
Vậy \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3+2+2+2=3+6=9\) (đpcm)
Ta có ;\(x+y+z=1\) \(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\left(1+1+1\right)^2=9\)(áp dụng bất đẳng thức Bunhiacopxki)
Vậy : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge9\)
Áp dụng bđt Cauchy schwarz:
=> 1/x+1/y+4/z+16/t >= [(1+1+2+4)^2] / x+y+z+t=8^2/(x+y+z+t)=64/1=64
=> đpcm.
Áp dụng BĐT Svac - xơ:
\(\frac{1}{x}+\frac{1}{y}+\frac{4}{z}+\frac{16}{t}\ge\frac{\left(1+1+2+4\right)^2}{x+y+z+t}=\frac{64}{1}=64\)
(Dấu "="\(\Leftrightarrow x=y=\frac{1}{22};z=\frac{2}{11};t=\frac{8}{11}\))
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Ta có \(1=a+b+c\ge3\sqrt[3]{abc}\)
\(\Leftrightarrow\frac{1}{3}\ge\sqrt[3]{abc}\)
Theo đề bài ta có
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}\)
\(\ge\frac{3\sqrt[3]{a^2b^2c^2}}{abc}=\frac{3}{\sqrt[3]{abc}}\ge9\)
\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow\)\(x+y+z=\frac{xy+yz+xz}{xyz}\)
\(\Leftrightarrow\)\(x+y+z=xy+yz+xz\) (vì xyz = 1 )
Ta có: \(\left(xyz-1\right)+\left(x+y+z\right)-\left(xy+yz+xz\right)=0\)
\(\Leftrightarrow\)\(\left(xyz-xy\right)-\left(xz-x\right)-\left(yz-y\right)+\left(z-1\right)=0\)
\(\Leftrightarrow\)\(xy\left(z-1\right)-x\left(z-1\right)-y\left(z-1\right)+\left(z-1\right)=0\)
\(\Leftrightarrow\)\(\left(z-1\right)\left(x-1\right)\left(y-1\right)=0\) (mk lm hơi tắt, thông cảm)
\(\Leftrightarrow\) \(x-1=0\) \(\Leftrightarrow\) \(x=1\)
hoặc \(y-1=0\) \(\Leftrightarrow\) \(y=1\)
hoặc \(z-1=0\) \(\Leftrightarrow\) \(z=1\)
Vậy....
a) \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\Leftrightarrow\frac{2+x^2+y^2}{\left(1+x^2\right)\left(1+y^2\right)}\ge\frac{2}{1+xy}\)
\(\Leftrightarrow\left(2+x^2+y^2\right)\left(1+xy\right)\ge2\left(1+x^2\right)\left(1+y^2\right)\)
\(\Leftrightarrow2+2xy+x^2+x^3y+y^2+y^3x\ge2\left(x^2+y^2+x^2y^2+1\right)\)
\(\Leftrightarrow x^3y+xy^3+2xy-x^2-y^2-2x^2y^2\ge0\)
\(\Leftrightarrow xy\left(x^2-2xy+y^2\right)-\left(x^2-2xy+y^2\right)\ge0\Leftrightarrow\left(xy-1\right)\left(x-y\right)^2\ge0\) (đúng)
Từ x+y+z=3 ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\frac{\Leftrightarrow xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)
Nhân chéo ta có:
\(\left(xy+yz+zx\right)\left(x+y+z\right)=xyz\)
\(\Leftrightarrow x^2y+xyz+x^2z+y^2x+y^2z+xyz+xyz+z^2y+z^2x=xyz\)
\(\Leftrightarrow x^2y+x^2z+y^2z+y^2x+z^2x+z^2y+2xyz=0\)
\(\Leftrightarrow\left(x^2y+x^2z+y^2x+xyz\right)+\left(y^2z+z^2x+z^2y+xyz\right)=0\)
\(\Leftrightarrow x\left(xy+xz+y^2+yz\right)+z\left(xy+xz+y^2+yz\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left(xy+xz+y^2+yz\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left[\left(xy+y^2\right)+\left(xz+yz\right)\right]=0\)
\(\Leftrightarrow\left(x+z\right)\left[y\left(x+y\right)+z\left(x+y\right)\right]=0\)
\(\Leftrightarrow\left(x+z\right)\left(y+z\right)\left(x+y\right)=0\)
Suy ra x+z=0 hoặc y+z=0 hoặc x+y=0
Với x+z=0 ta đc y=3
Với y+z=0 ta đc x=3
Với x+y=0 ta đc z=3
Từ đó suy ra đccm
Với x,y,z > 0
Xét : (1/x + 1/y + 1/z).(x+y+z)
>=3 \(\sqrt[3]{\frac{1}{xyz}}\). 3\(\sqrt[3]{xyz}\) = 9
=> 1/x + 1/y + 1/z >= 9/x+y+z = 9/1 = 9
=> ĐPCM
Dấu "=" xảy ra <=> x=y=z=1/3
Tk mk nha
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
\(\Leftrightarrow3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge9\)(đúng với a, b, c dương)
Áp dụng BĐT trên ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}=9\)