Cho x,y,z>0,x+y+z\(\ge\)12.CMR \(\frac{x}{\sqrt{y}}\)+\(\frac{y}{\sqrt{z}}\)+\(\frac{z}{\sqrt{x}}\)\(\ge\)6
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T
20 tháng 11 2019
Ta có: \(\frac{1}{2}.2x\left(1-x\right)\left(1-x\right)\le\frac{1}{2}\left[\frac{2x+1-x+1-x}{3}\right]^3=\frac{4}{27}\)
\(\Rightarrow\sqrt{x}\left(1-x\right)\le\frac{2\sqrt{3}}{9}\Rightarrow\frac{1}{\sqrt{x}\left(1-x\right)}\ge\frac{9}{2\sqrt{3}}\)
\(\Rightarrow\frac{\sqrt{x}}{1-x}\ge\frac{3\sqrt{3}}{2}x\). Thiết lập tương tự hai BĐT còn lại và cộng theo vế thu được đpcm.
TN
12 tháng 10 2016
\(A=\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\)
Áp dụng Bđt MIncopxki ta có:
\(A\ge\sqrt{\left(x+y+\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\)
\(\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)
\(\ge\sqrt{\left(x+y+z\right)^2+\frac{1}{\left(x+y+z\right)^2}+\frac{80}{\left(x+y+z\right)^2}}\)
\(\ge\sqrt{2+80}=\sqrt{82}\)
Dấu = khi \(x=y=z=\frac{1}{3}\)
N
2
1 tháng 1 2020
\(\left(1.x+9.\frac{1}{y}\right)^2\le\left(1^2+9^2\right)\left(x^2+\frac{1}{y^2}\right)\Rightarrow\sqrt{x^2+\frac{1}{y^2}}\)
\(\ge\frac{1}{\sqrt{82}}\left(x+\frac{9}{y}\right)\)
\(TT:\sqrt{y^2+\frac{1}{z^2}}\ge\frac{1}{\sqrt{82}}\left(x+\frac{9}{z}\right);\sqrt{z^2+\frac{1}{x^2}}\ge\frac{1}{\sqrt{82}}\left(z+\frac{9}{x}\right)\)
\(S\ge\frac{1}{\sqrt{82}}\left(x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\right)\)
\(\ge\frac{1}{\sqrt{82}}\left(x+y+z+\frac{81}{x+y+z}\right)\)
\(=\frac{1}{\sqrt{82}}\left[\left(x+y+z+\frac{1}{x+y+z}\right)+\frac{80}{x+y+z}\right]\ge\sqrt{82}\)
Ta có: \(\sqrt{y}\le\frac{y+4}{4}\) (bđt cosi) => \(\frac{x}{\sqrt{y}}\ge\frac{4x}{y+4}=\frac{4x^2}{xy+4x}\)
CMTT: \(\frac{y}{\sqrt{z}}\ge\frac{4y}{z+4}=\frac{4y^2}{yz+4y}\)
\(\frac{z}{\sqrt{x}}\ge\frac{4z}{x+4}=\frac{4z^2}{xz+4z}\)
=>A = \(\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{z}}+\frac{z}{\sqrt{x}}\ge\frac{4x^2}{xy+4x}+\frac{4y^2}{yz+4y}+\frac{4z^2}{xz+4z}=4\left(\frac{x^2}{xy+4x}+\frac{y^2}{yz+4y}+\frac{z^2}{xz+4z}\right)\)
=> A \(\ge4\cdot\frac{\left(x+y+z\right)^2}{yz+xz+xy+4\left(x+y+z\right)}\)(bđt svacxo: \(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3}\ge\frac{\left(x_1+x_2+x_3\right)^2}{y_1+y_2+y_3}\))
<=> A \(\ge4.\frac{\left(x+y+z\right)^2}{\frac{\left(x+y+z\right)^2}{3}+4\left(x+y+z\right)}\)(bđt: ab + bc + ac \(\le\)(a + b + c)2/3
<=> A \(\ge\frac{12\left(x+y+z\right)^2}{\left(x+y+z\right)\left(x+y+z+12\right)}=\frac{12\left(x+y+z\right)}{x+y+z+12}=\frac{12\left(x+y+z+12\right)}{x+y+z+12}-\frac{144}{x+y+z+12}\)
A \(\ge12-\frac{144}{12+12}=12-6=6\)