1) Chứng minh rằng nếu: xyz=1 thì \(\frac{1}{1+x+xy}\)+\(\frac{1}{1+y+yz}\)+\(\frac{1}{1+z+zx}\)=1
2) Cho \(\frac{x^2}{x+y}\)+\(\frac{y^2}{y+z}\)+\(\frac{z^2}{z+x}\)=2017. Tính: \(\frac{x^2}{x+y}\)+\(\frac{z^2}{y+z}\)+\(\frac{x^2}{z+x}\)
cho x,y,z>0 với xy+yz+zx=3
Chứng minh rằng \(\frac{1}{1+x^2\left(y+z\right)}+\frac{1}{1+y^2\left(x+z\right)}+\frac{1}{1+z^2\left(y+x\right)}\le\frac{1}{xyz}\)
Chứng minh rằng:
a, nếu x+y=1 thì \(\frac{x}{y^3-1}+\frac{y}{x^3-1}+\frac{2\left(xy-2\right)}{x^2y^2+3}=0\)
b, nếu x,y,z khác -1 thì\(\frac{xy+2x+1}{xy+x+y+1}+\frac{yz+2y+1}{yz+z+y+1}+\frac{zx+2z+1}{zx+z+x+1}=3\)
c, Cho x,y,z đôi một khác nhau thỏa mãn\(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}=0\) thì\(\frac{x}{\left(y-z\right)^2}+\frac{y}{\left(z-x\right)^2}+\frac{z}{\left(x-y\right)^2}=0\)
Chứng minh rằng: \(\frac{x-y}{1+xy}+\frac{y-z}{1+yz}+\frac{z-x}{1+zx}=\frac{x-y}{1+xy}\cdot\frac{y-z}{1+yz}\cdot\frac{z-x}{1+zx}\)
Cho x,y,z > 0 thỏa mãn xy + yz +zx = 1.Chứng minh
\(\frac{x-y}{z^2+1}\)+\(\frac{y-z}{x^2+1}\)+\(\frac{z-x}{y^2+1}\)=0
\(\dfrac{x-y}{z^2+1}=\dfrac{x-y}{z^2+xy+yz+zx}=\dfrac{x-y}{z\left(z+y\right)+x\left(z+y\right)}=\dfrac{x-y}{\left(x+z\right)\left(z+y\right)}\)
Tương tự: \(\dfrac{y-z}{x^2+1}=\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}\);\(\dfrac{z-x}{y^2+1}=\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
Cộng vế với vế \(\Rightarrow VT=\dfrac{x-y}{\left(x+z\right)\left(y+z\right)}+\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}+\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\dfrac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(đpcm)
Cho \(\hept{\begin{cases}x,y,z>0\\xy+yz+zx=1\end{cases}}\). Chứng minh rằng:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ge3+\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x^2}}+\sqrt{\frac{\left(y+z\right)\left(y+x\right)}{y^2}}+\sqrt{\frac{\left(z+x\right)\left(z+y\right)}{z^2}}\)
1111111111111111111
\(VT=\Sigma\frac{xy+yz+zx}{xy}=3+\Sigma\frac{z\left(x+y\right)}{xy}\)
Đến đây để ý \(\frac{1}{2}\left[\frac{z\left(x+y\right)}{xy}+\frac{y\left(z+x\right)}{zx}\right]\ge\sqrt{\frac{\left(z+x\right)\left(x+y\right)}{x^2}}\left(\text{AM - GM}\right)\)
Là xong.
1.Giải hệ pt
1)\(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\\xy+yz+zx=3\\\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}=x\end{cases}}\)
2)\(\hept{\begin{cases}xy+yz+zx=3\\\left(x+y\right)\left(y+z\right)=\sqrt{3}z\left(1+y^2\right)\\\left(y+z\right)\left(z+x\right)=\sqrt{3}x\left(1+z^2\right)\end{cases}}\)
3)\(\hept{\begin{cases}xy+yz+zx=3\\1+x^2\left(y+z\right)+xyz=4y\\1+y^2\left(z+x\right)+xyz=4z\end{cases}}\)
Cho các số dương x,y,z thỏa mãn: xy + yz + zx = 3xyz. Chứng minh rằng
\(\frac{x^3}{x^2+z}+\frac{y^3}{y^2+x}+\frac{z^3}{z^2+y}\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Theo GT : \(xy+yz+xz=3xyz\Rightarrow\frac{xy+yz+xz}{xyz}=3\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
\(\frac{x^3}{x^2+z}=\frac{x\left(x^2+z\right)}{x^2+z}-\frac{xz}{x^2+z}=x-\frac{xz}{x^2+z}\ge x-\frac{xz}{2x\sqrt{z}}=x-\frac{\sqrt{z}}{2}\)
Tương tự , ta có : \(\frac{y^3}{y^2+x}\ge y-\frac{\sqrt{x}}{2}\) ; \(\frac{z^3}{z^2+y}\ge z-\frac{\sqrt{y}}{2}\)
\(\Rightarrow\frac{x^3}{x^2+z}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+y}\ge x+y+z-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\)
Vì x ; y ; z dương , áp dụng BĐT Cô - si , ta có :
\(x+1\ge2\sqrt{x};y+1\ge2\sqrt{y};z+1\ge2\sqrt{z}\)
\(\Rightarrow x+y+z+3\ge2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)
=> \(\frac{x+y+z+3}{2}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}\) => BĐT được c/m
Tiếp tục AD BĐT Cô - si , ta có :
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{xyz}}=9\)
\(\Rightarrow x+y+z\ge\frac{9}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac{9}{3}=3\) => BĐT được c/m
Có : \(\frac{x^3}{x^2+z}+\frac{y^3}{y^2+x}+\frac{z^3}{z^2+y}\ge x+y+z-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\ge x+y+z-\frac{x+y+z+3}{4}=\frac{3x+3y+3z-3}{2}\ge\frac{3.3-3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=z=1\)
Vậy ...
Cho x, y, z > 0 thoả mãn: \(xy+yz+zx=3xyz\). Chứng minh rằng: \(\frac{x^3}{z+x^2}+\frac{y^3}{x+y^2}+\frac{z^3}{y+z^2}\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Lời giải:
Áp dụng BĐT AM-GM ta có:
\(\text{VT}=x-\frac{x}{x^2+z}+y-\frac{y}{y^2+x}+z-\frac{z}{z^2+y}=(x+y+z)-\left(\frac{x}{x^2+z}+\frac{y}{y^2+x}+\frac{z}{z^2+y}\right)\)
\(\geq (x+y+z)-\left(\frac{x}{2\sqrt{x^2z}}+\frac{y}{2\sqrt{y^2x}}+\frac{z}{2\sqrt{z^2y}}\right)=(x+y+z)-\frac{1}{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)(1)\)
Từ giả thiết \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
Cauchy-Schwarz:
\(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3(2)\)
\(\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2\leq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(1+1+1)=9\)
\(\Rightarrow \left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)\leq 3(3)\)
Từ \((1);(2);(3)\Rightarrow \text{VT}\geq 3-\frac{1}{2}.3=\frac{3}{2}\)
Mặt khác: \(\text{VP}=\frac{1}{2}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{2}\)
Do đó \(\text{VT}\geq \text{VP}\) (đpcm)
Dấu "=" xảy ra khi $x=y=z=1$
Cho x,y,z>0 thỏa mãn xy+yz+zx=1. Chứng minh \(\frac{x}{x^2-yz+3}+\frac{y}{y^2-zx+3}+\frac{z}{z^2-xy+3}\ge\frac{1}{x+y+z}\)