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2 tháng 5 2018

viết lại bt P đi

28 tháng 1 2019

CMR : \(\frac{x}{yz+1}+\frac{y}{xz+1}+\frac{z}{xy+1}\le2;\left(0\le x\le y\le z\le1\right)\)

Ta có : \(\frac{x}{yz+1}+\frac{y}{xz+1}+\frac{z}{xy+1}\le\frac{x}{xy+1}+\frac{y}{xy+1}+\frac{z}{xy+1}=\frac{x+y+z}{xy+1}\left(1\right)\)

Ta lại có : \(0\le x\le1;0\le y\le1\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\ge0\)

\(\Leftrightarrow xy-x-y+1\ge0\)

\(\Leftrightarrow xy+1\ge x+y\left(2\right)\)

Thay (2) và (1) được : \(\frac{x+y+z}{xy+1}\le\frac{xy+1+2}{xy+1}\le\frac{2\left(xy+1\right)}{xy+1}=2\)

16 tháng 5 2020

Vì \(0\le x\le y\le z\le1\Rightarrow x-1\le0;y-1\le0\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)\ge0\Rightarrow xy+1\ge x+y\Rightarrow\frac{1}{xy+1}\le\frac{1}{x+y}\Rightarrow\frac{z}{xy+1}\le\frac{z}{x+y}\left(1\right)\)

Cmtt: \(\hept{\begin{cases}\frac{x}{yz+1}\le\frac{x}{y+z}\left(2\right)\\\frac{y}{xz+1}\le\frac{y}{x+z}\left(3\right)\end{cases}}\)

Từ (1), (2), (3) ta có:

\(\frac{x}{yz+1}+\frac{y}{xz+1}+\frac{z}{xy+1}\le\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\left(4\right)\)

Mà \(\frac{x}{y+z}\le\frac{x+z}{x+y+z}\Rightarrow\frac{x}{y+z}\le\frac{2x}{x+y+z}\)

Cmtt: \(\hept{\begin{cases}\frac{y}{x+z}\le\frac{2y}{x+y+z}\\\frac{z}{x+y}\le\frac{2z}{x+y+z}\end{cases}}\)

\(\Rightarrow\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\le\frac{2\left(x+y+z\right)}{x+y+z}\le2\left(5\right)\)

Từ (4), (5) => đpcm

5 tháng 4 2017

Áp dụng BĐT AM-GM ta có: 

\(\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x\ge2\sqrt{9^x}=2\cdot3^x\)

\(\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\ge2\sqrt{25^x}=2\cdot5^x\)

\(\left(\frac{20}{3}\right)^x+\left(\frac{12}{5}\right)^x\ge2\sqrt{16^x}=2\cdot4^x\)

Cộng theo vế 3 BĐT trên ta có: 

\(2\left[\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\right]\ge2\left(3^x+4^x+5^x\right)\)

\(\Rightarrow\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\ge3^x+4^x+5^x\)

28 tháng 2 2019

Ta có: \(z^2=2\left(xz+yz-xy\right)=2xz+2yz-2xy\)

Xét:

\(x^2+\left(x-z\right)^2=x^2+z^2-z^2+\left(x-z\right)^2\)\(=\left(x-z\right)^2+2xz-\left(2xz+2yz-2xy\right)+\left(x-z\right)^2\)

\(=\left(x-z\right)^2+2xy-2yz+\left(x-z\right)^2=\left(x-z\right)^2+2y\left(x-z\right)+\left(x-z\right)^2\)

\(=\left(x-z\right)\left(x-z+2y+x-z\right)=\left(x-z\right)\left(2x+2y-2z\right)\)                                    (1)

Xét:

\(y^2+\left(y-z\right)^2=y^2+z^2-z^2+\left(y-z\right)^2\)\(=\left(y-z\right)^2+2yz-\left(2xz+2yz-2xy\right)\)

\(=\left(y-z\right)^2+2xy-2xz+\left(y-z\right)^2=\left(y-z\right)^2+2x\left(y-z\right)+\left(y-z\right)^2\)

\(=\left(y-z\right)\left(y-z+2x+y-z\right)=\left(y-z\right)\left(2x+2y-2z\right)\)                                      (2)

Từ (1); (2) => \(\frac{x^2+\left(x-z\right)^2}{y^2+\left(y-z\right)^2}=\frac{\left(x-z\right)\left(2x+2y-2z\right)}{\left(y-z\right)\left(2x+2y-2z\right)}=\frac{x-z}{y-z}\) \(\left(ĐPCM\right)\)