giả sử \(x\ge y\ge z\ge0\)chứng minh :
\(\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2\)
Cho \(x\ge y\ge z\ge0\). Chứng minh BĐT sau
a/ \(xy^3+yz^3+zx^3\ge xz^3+zy^3+yx^3\)
b/ \(\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge\dfrac{x^2z}{y}+\dfrac{y^2x}{z}+\dfrac{z^2y}{x}\)
a) BĐT \(\Leftrightarrow\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)\ge0\)
suy ra sai đề
b) BĐT \(\Leftrightarrow\dfrac{\left(x-y\right)\left(y-z\right)\left(x-z\right)\left(xy+yz+xz\right)}{xyz}\ge0\) ( đúng vì \(x\ge y\ge z>0\))
Cho \(x\ge y\ge z>0\)
CMR : \(\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2\)
Cho x, y, z>0. Chứng minh rằng:
\(\dfrac{x}{x+2y+3z}+\dfrac{y}{y+2z+3x}+\dfrac{z}{z+2x+3y}\ge\dfrac{1}{2}\)
\(VT=\dfrac{x^2}{x^2+2xy+3zx}+\dfrac{y^2}{y^2+2yz+3xy}+\dfrac{z^2}{z^2+2zx+3yz}\)
\(VT\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+5xy+5yz+5zx}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+3\left(xy+yz+zx\right)}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(x+y+z\right)^2}=\dfrac{1}{2}\)
cho x,y,z là các số thực dương thỏa mãn \(x^2+y^2+z^2\ge\dfrac{1}{3}\)
chứng minh \(\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\ge\dfrac{1}{30}\)
đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)
\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)
BBDT AM-GM
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)
vì \(x^2+y^2\ge2xy\)
\(y^2+z^2\ge2yz\)
\(x^2+z^2\ge2xz\)
\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)
\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)
dấu"=" xảy ra<=>x=y=z=1/3
Cho các số dương x,y,z và \(x^2+y^2+z^2=1\).Chứng minh rằng:\(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}\ge\dfrac{1}{3}\)
\(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2xz}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{xz+2yz}\)
\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)
Cho các số dương x;y;z thỏa mãn : x\(\ge\)y\(\ge\)z .Cm rằng :
\(\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2\)
Đề bài:Cho x,y,z dương thỏa mãn \(x\geq y\geq z>0\). CMR
\(\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}\geq x^2+y^2+z^2\)
Giải
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\right)\left(\dfrac{x^2z}{y}+\dfrac{y^2x}{z}+\dfrac{z^2y}{x}\right)\ge\left(x^2+y^2+z^2\right)^2\)
Vậy ta cần chứng minh \(\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge\dfrac{x^2z}{y}+\dfrac{y^2x}{z}+\dfrac{z^2y}{x}\)
Thật vậy ta có: \(\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}-\dfrac{x^2z}{y}+\dfrac{y^2x}{z}+\dfrac{z^2y}{x}\ge0\)
\(\Leftrightarrow\dfrac{\left(xy+yz+xz\right)\left(x-y\right)\left(y-z\right)\left(x-z\right)}{xyz}\ge0\) (luôn đúng)
Cho các số dương x, y, z thỏa mãn điều kiện \(x^2+y^2+z^2=1\).CM \(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}\ge\dfrac{1}{3}\)
mong mọi nguòi giúp thank you
Ta có: \(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2zx}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{zx+2yz}\)
\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+2zx+yz+2xy+zx+2yz}=\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\)
Mà ta lại có: \(xy+yz+zx\le x^2+y^2+z^2\)
\(\Rightarrow\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1^2}{3.1}=\dfrac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{\sqrt{3}}\)
Cho các số x, y, z dương thỏa mãn: \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=3\)
Cmr: \(\dfrac{1}{\left(2x+y+z\right)^2}+\dfrac{1}{\left(2y+z+x\right)^2}+\dfrac{1}{\left(2z+x+y\right)^2}\ge\dfrac{3}{16}\)
Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!
Cho \(x,y,z\ge0\) chứng minh:
\(\dfrac{x+y}{\left(x-y\right)^2}+\dfrac{z+y}{\left(y-z\right)^2}+\dfrac{x+z}{\left(x-z\right)^2}\ge\dfrac{9}{x+y+z}\)