cho x>y>z CMR x^4(y-z)+y^4(z-x)+z^4(x-y)>0
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TL
0
NA
1
S
8 tháng 10 2018
Sửa đề z^4(z-y) thành z^4(x-y)
Đặt \(A=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)
\(=x^4\left(y-x+x-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)
\(=-x^4\left(x-y\right)+x^4\left(x-z\right)-y^4\left(x-z\right)+z^4\left(x-y\right)\)
\(=\left(x-y\right)\left(z^4-x^4\right)+\left(x-z\right)\left(x^4-y^4\right)\)
\(=\left(x-y\right)\left(z^2+x^2\right)\left(z^2-x^2\right)+\left(x-z\right)\left(x^2+y^2\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(z^2+x^2\right)\left(x+z\right)\left(z-x\right)+\left(x-z\right)\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(z-x\right)\left[\left(z^2+x^2\right)\left(x+z\right)-\left(x^2+y^2\right)\left(x+y\right)\right]\)
\(=\left(x-y\right)\left(z-x\right)\left(xz^2+z^3+x^3+x^2z-x^3-x^2y-xy^2-y^3\right)\)
\(=\left(x-y\right)\left(z-x\right)\left[x^2\left(z-y\right)+x\left(z^2-y^2\right)+\left(z^3-y^3\right)\right]\)
\(=\left(x-y\right)\left(z-x\right)\left(z-y\right)\left[x^2+x\left(z+y\right)+\left(z^2+yz+y^2\right)\right]\)
\(=\left(x-y\right)\left(x-z\right)\left(y-z\right)\left(x^2+xz+xy+z^2+yz+y^2\right)\)
\(=\frac{1}{2}\left(x-y\right)\left(x-z\right)\left(y-z\right)\left(2x^2+2y^2+2z^2+2xy+2yz+2xz\right)\)
\(=\frac{1}{2}\left(x-y\right)\left(x-z\right)\left(y-z\right)\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\right]\)
Vì \(x>y>z\Rightarrow\hept{\begin{cases}x-y>0\\x-z>0\\y-z>0\end{cases}}\) và \(\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\ge0\)
=>....
VS
1
KK
20 tháng 1 2017
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}x+y\ge2\sqrt{xy}\\\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{xy}}\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\sqrt{xy.\frac{1}{xy}}\)
\(\Rightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\) ( đpcm )
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}x+y+z\ge3\sqrt{xyz}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt{\frac{1}{xyz}}\end{matrix}\right.\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\sqrt{xyz.\frac{1}{xyz}}\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) ( đpcm )
VD
0
H
1
11 tháng 12 2016
Ta có:
\(\frac{x}{x+1}=1-\frac{1}{x+1}\)
\(\frac{y}{y+1}=1-\frac{y}{y+1}\)
\(\frac{z}{z+4}=1-\frac{4}{z+4}\)
\(\Rightarrow\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+4}=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{4}{z+4}\right)\)
\(\le\left[3-\left(\frac{4}{x+y+2}+\frac{4}{z+4}\right)\right]\le\left(3-\frac{16}{x+y+z+6}\right)=3-\frac{16}{6}=\frac{1}{3}\)
\(x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)
Ta có: \(x^4\ge0;y^4\ge0;z^4\ge0\)
\(x>y\Rightarrow x^4>y^4\)
\(y>z\Rightarrow y-z>0\)
\(x>z\Rightarrow z-x< 0\)
\(\Rightarrow y-z>z-x\)
\(\Rightarrow x^4\left(y-z\right)+y^4\left(z-x\right)>0\)
\(x>y\Rightarrow x-y>0\)
Vậy: \(x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)>0\)