cho x,y,z là ba số dương thỏa mãn điều kiện
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=2\)
Chứng minh \(xyz\le\frac{1}{8}\)
K
Khách
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27 tháng 9 2018
\(\frac{1}{x+1}=1-\frac{1}{y+1}+1-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)
Tương tụ co:
\(\hept{\begin{cases}\frac{1}{y+1}\ge2\sqrt{\frac{zx}{\left(z+1\right)\left(x+1\right)}}\\\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\end{cases}}\)
\(\Rightarrow\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\frac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(\Leftrightarrow xyz\le\frac{1}{8}\)
KN
9 tháng 12 2020
Ta có: \(x+y+z=xyz\Rightarrow x=\frac{x+y+z}{yz}\Rightarrow x^2=\frac{x^2+xy+xz}{yz}\Rightarrow x^2+1=\frac{\left(x+y\right)\left(x+z\right)}{yz}\)\(\Rightarrow\sqrt{x^2+1}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{yz}}\le\frac{\frac{x+y}{y}+\frac{x+z}{z}}{2}=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}\le\frac{2+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự: \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Cộng theo vế ba bất đẳng thức trên, ta được: \(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3.\frac{xy+yz+zx}{xyz}\)\(\le3.\frac{\frac{\left(x+y+z\right)^2}{3}}{xyz}=\frac{\left(x+y+z\right)^2}{xyz}=\frac{\left(xyz\right)^2}{xyz}=xyz\)
Đẳng thức xảy ra khi \(x=y=z=\sqrt{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 9 2020
Đặt \(\left(x;y;z\right)=\left(2a^2;2b^2;2c^2\right)\Rightarrow abc=1\)
\(VT=\frac{1}{4a^2+2b^2+6}+\frac{1}{4b^2+2c^2+6}+\frac{1}{4c^2+2a^2+6}\)
\(VT=\frac{1}{\left(2a^2+2\right)+\left(2a^2+2b^2\right)+4}+\frac{1}{\left(2b^2+2\right)+\left(2b^2+2c^2\right)+4}+\frac{1}{\left(2c^2+2\right)+\left(2c^2+2a^2\right)+4}\)
\(VT\le\frac{1}{4a+4ab+4}+\frac{1}{4b+4bc+4}+\frac{1}{4c+4ca+4}=\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\) hay \(x=y=z=2\)
26 tháng 4 2020
\(\frac{1}{x^2+2y^2+3}+\frac{1}{y^2+2z^2+3}+\frac{1}{z^2+2x^2+3}\)
= \(\frac{1}{x^2+y^2+y^2+1+2}+\frac{1}{y^2+z^2+z^2+1+2}+\frac{1}{z^2+x^2+x^2+1+2}\)
\(\le\frac{1}{2xy+2y+2}+\frac{1}{2yz+2z+2}+\frac{1}{2zx+2x+2}\)
= \(\frac{1}{2}\left(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}\right)\)
= \(\frac{1}{2}\left(\frac{zx}{xyzx+yzx+zx}+\frac{x}{yzx+zx+x}+\frac{1}{zx+x+1}\right)\)
= \(\frac{1}{2}\left(\frac{zx}{x+1+zx}+\frac{x}{1+zx+x}+\frac{1}{zx+x+1}\right)\)
= 1/2
Dấu "=" xảy ra <=> x = y =z =1
26 tháng 4 2020
Áp dụng BĐT AM-GM ta có:\(\hept{\begin{cases}x^2+y^2\ge2xy\\y^2+1\ge2y\end{cases}\Rightarrow\frac{1}{x^2+2y^2+3}\le\frac{1}{2xy+2y+2}}\)
Tương tự ta cũng có
\(\frac{1}{y^2+2x^2+3}\le\frac{1}{2yz+2z+2};\frac{1}{z^2+2x^2+3}\le\frac{1}{2xz+2x+2}\)
Do đó ta có:\(VT\le\frac{1}{2}\left(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}\right)\)
Mặt khác, do xyz=1 nên ta có:
\(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}=\frac{1}{xy+y+1}+\frac{y}{xy+y+1}+\frac{xy}{xy+y+1}\)
\(=\frac{xy+y+1}{xy+y+1}=1\)
\(\Rightarrow VT\le\frac{1}{2}\). Dấu "=" xảy ra <=> x=y=z=1
LM
20 tháng 7 2017
Áp dụng BĐT AM-GM ta có: \(\frac{1}{x+1}=\)\(1-\frac{1}{y+1}+1-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{ỹz}{\left(y+1\right)\left(z+1\right)}}\)
Tương tự ta cũng có: \(\frac{1}{y+1}\ge2\sqrt{\frac{xz}{\left(x+1\right)\left(z+1\right)}};\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)
Nhân từng vế của ba BĐT trên ta được:
\(\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge8\frac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\Rightarrow xyz\le\frac{1}{8}\)
1 tháng 11 2019
Ta có :
\(\frac{1+\sqrt{1+x^2}}{x}=\frac{2+\sqrt{4\left(1+x^2\right)}}{2x}\le\frac{2+\frac{4+1+x^2}{2}}{2x}=\frac{9+x^2}{4x}\)
tương tự : \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{9+y^2}{4y}\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{9+z^2}{4z}\)
\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le\frac{\left(9+x^2\right)yz+\left(9+y^2\right)xz+\left(9+z^2\right)xy}{4xyz}\)
\(=\frac{9\left(xy+yz+xz\right)+xyz\left(x+y+z\right)}{4xyz}\le\frac{9\frac{\left(x+y+z\right)^2}{3}+\left(xyz\right)^2}{4xyz}=\frac{4\left(xyz\right)^2}{4xyz}=xyz\)
Dấu " = " xảy ra khi x = y = z = \(\sqrt{3}\)
Ta có: \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=2\)
\(\Rightarrow\frac{1}{x+1}=2-\frac{1}{y+1}-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}=\frac{2\sqrt{yz}}{\sqrt{\left(y+1\right)\left(z+1\right)}}\) (1)
(Vì x;y;z dương nên áp dụng BĐT Cô-si)
Chưng minh tương tự ta có: \(\frac{1}{y+1}\ge2\sqrt{\frac{xz}{\left(x+1\right)\left(z+1\right)}}=\frac{2\sqrt{xz}}{\sqrt{\left(x+1\right)\left(z+1\right)}}\) (2)
\(\frac{1}{z+1}\ge\frac{2\sqrt{xy}}{\sqrt{\left(x+1\right)\left(y+1\right)}}\) (3)
Nhân (1) với (2) với (3) ta có:
giải tiếp
\(\frac{1}{x+1}.\frac{1}{y+1}.\frac{1}{z+1}=\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\frac{8\sqrt{\left(xyz\right)^2}}{\sqrt{\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2}}\)
Với x;y;z > 0 nên \(1\ge\frac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\Leftrightarrow1\ge8xyz\Leftrightarrow xyz\le\frac{1}{8}\)
Vậy ....