Cho x,y,z > 2 và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\). CMR: \(\left(x-2\right)\left(y-2\right)\left(z-2\right)\le1\)
Cho x,y,z>2 tm: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\). CMR: \(\left(x-2\right)\left(y-2\right)\left(z-2\right)\le1\)
Đặt \(\hept{\begin{cases}a=x-2\\b=y-2\\c=z-2\end{cases}}\left(a,b,c>0\right)\)
Lúc đó giả thiết được viết lại thành \(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1\)và ta cần chứng minh \(abc\le1\)
Ta có: \(\frac{1}{a+2}=1-\frac{1}{b+2}-\frac{1}{c+2}=\frac{1}{2}-\frac{1}{b+2}+\frac{1}{2}-\frac{1}{c+2}\)
\(=\frac{b}{2\left(b+2\right)}+\frac{c}{2\left(c+2\right)}\ge2\sqrt{\frac{bc}{4\left(b+2\right)\left(c+2\right)}}\)(Theo bất đẳng thức Cauchy cho 2 số dương) (1)
Hoàn toàn tương tự: \(\frac{1}{b+2}\ge2\sqrt{\frac{ca}{4\left(c+2\right)\left(a+2\right)}}\)(2) ; \(\frac{1}{c+2}\ge2\sqrt{\frac{ab}{4\left(a+2\right)\left(b+2\right)}}\)(3)
Nhân theo vế 3 bất đẳng thức (1), (2), (3), ta được:
\(\frac{1}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\ge\frac{abc}{\sqrt{\left(a+2\right)^2\left(b+2\right)^2\left(c+2\right)^2}}\)
\(\Leftrightarrow\frac{1}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\ge\frac{abc}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\Leftrightarrow abc\le1\)(đpcm)
Đẳng thức xảy ra khi \(x=y=z=3\)
CHo x,y,z>2 thỏa mãn \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\) CMR: \(\left(x-2\right)\left(y-2\right)\left(z-2\right)\le1\)
Cho x,y,z>2 thỏa mãn \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\) CMR : \(\left(x-2\right)\left(y-2\right)\left(z-2\right)\le1\)
Đặt a=x-2; b=y-2; c=z-2. Phải chứng minh abc =<1
Thật vậy, từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)ta có:
\(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1\)
Theo BĐT Cauchy ta có:
\(\frac{1}{a+2}=\left(\frac{1}{2}-\frac{1}{b+2}\right)+\left(\frac{1}{2}-\frac{1}{c+2}\right)=\frac{1}{2}\left(\frac{b}{b+2}+\frac{c}{c+2}\right)\ge\sqrt{\frac{bc}{\left(b+2\right)\left(c+2\right)}}\left(1\right)\)
tương tự \(\hept{\begin{cases}\frac{1}{b+2}\ge\sqrt{\frac{ac}{\left(a+2\right)\left(c+2\right)}}\left(2\right)\\\frac{1}{c+2}\ge\sqrt{\frac{ab}{\left(a+2\right)\left(b+2\right)}}\left(3\right)\end{cases}}\)
Nhân các vế của (1)(2)(3) ta được đpcm
Dấu "=" xảy ra <=> a=b=c hay x=y=z=3
Cho x,y,z>0 và xy+yz+zx=1
a, tính giá trị biểu thức:
\(P=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
b, CMR:
\(\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}=\frac{2xy}{\sqrt{\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)}}\)
Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+y\right)\left(x+z\right)\)
Tương tự \(1+y^2=\left(x+y\right)\left(y+z\right)\)
\(1+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào A ta được
\(P=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
=2(xy+xz+yz)=2
\(b,VT=VP\)
\(\Leftrightarrow\frac{x}{xy+yz+zx+x^2}+\frac{y}{xy+yz+zx+y^2}+\frac{z}{xy+yz+zx+z^2}\)
\(=\frac{2xyz}{\sqrt{\left(xy+yz+zx+x^2\right)\left(xy+yz+zx+y^2\right)\left(xy+yz+zx+z^2\right)}}\)
\(\Leftrightarrow\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(x+y\right)\left(y+z\right)}+\frac{z}{\left(x+z\right)\left(y+z\right)}\)
\(=\frac{2xyz}{\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(y+x\right)\left(z+x\right)\left(y+z\right)}}\)
\(\Leftrightarrow\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(\Leftrightarrow xy+xz+xy+yz+xz+yz=2xyz\)
\(\Leftrightarrow2=2xyz\)
\(\Leftrightarrow xyz=1\)
Đù =)))
Cho x,y,z thỏa mãn 0<x,y,z<hoặc = 1 và x+y+z=2 CMR \(\frac{\left(x-1\right)^2}{z}+\frac{\left(y-1\right)^2}{x}+\frac{\left(z-1\right)^2}{y}\ge\frac{1}{2}\)
Cho x,y,z dương thỏa mãn xyz=1.CMR :
1) A\(=\frac{1}{x^2+x+1}+\frac{1}{y^2+y+1}+\frac{1}{z^2+z+1}\ge1\)
2) B\(=\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(y+1\right)\left(y+2\right)}+\frac{1}{\left(z+1\right)\left(z+2\right)}\ge\frac{1}{2}\)
cau a la bdt vas
con cau b la van dung he qua cua bdt vas
\(0< x,y,z\le1;x+y+z=2.\) tìm gtnn:
\(\frac{\left(x-1\right)^2}{z}+\frac{\left(y-1\right)^2}{x}+\frac{\left(z-1\right)^2}{y}\)
ap dung bdt cauchy schwarz ta co
\(\frac{\left(x-1\right)^2}{z}+\frac{\left(y-1\right)^2}{x}+\frac{\left(z-1\right)^2}{y}>=\frac{\left(x-1+z-1+y-1\right)^2}{x+y+z}=\frac{1}{2}\)
vay min=1/2
thực hiện phép tính
a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)
b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)
c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
cho x,y,z khác 0 và\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=x+y+z\)
cmr \(y\left(y^2-yz\right)\left(1-xz\right)=x\left(1-yz\right)\left(y^2-xz\right)\)