Xét pt thứ nhất \(\Leftrightarrow\sqrt{4x^2+4xy+4y^2}+\sqrt{4y^2+4yz+4z^2}+\sqrt{4z^2+4xz+4x^2}=6\)
\(\Leftrightarrow\sqrt{3\left(x-y\right)^2+\left(x+y\right)^2}+\sqrt{3\left(y-z\right)^2+\left(y+z\right)^2}+\sqrt{3\left(z-x\right)^2+\left(z+x\right)^2}=6\)
\(VT\ge x+y+y+z+z+x=2\left(x+y+z\right)=6\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
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)\(\hept{\begin{cases}xy+yz+zx=3\\1+x^2\left(y+z\right)+xyz=4y\\1+y^2\left(z+x\right)+xyz=4z\end{cases}}\)
Giải hệ PT sau: \(\hept{\begin{cases}3\sqrt{x}+2\sqrt{y}+\sqrt{z}=\frac{\sqrt{xyz}}{6}\\6\sqrt{xy}+2\sqrt{yz}+3\sqrt{zx}=108+18\sqrt{x+4}+12\sqrt{y+9}+6\sqrt{z+36}\end{cases}}\)
Cho x,y,z dương. Chứng minh \(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)\)
đặt \(A=\frac{\sqrt{yz}}{x+3\sqrt{yz}}+\frac{\sqrt{zx}}{y+3\sqrt{zx}}+\frac{\sqrt{xy}}{z+3\sqrt{xy}}\)
\(\Rightarrow1-3A=\frac{x}{x+3\sqrt{yz}}+\frac{y}{y+3\sqrt{zx}}+\frac{z}{z+3\sqrt{xy}}\)
\(\ge\frac{x}{x+\frac{3}{2}\left(y+z\right)}+\frac{y}{y+\frac{3}{2}\left(z+x\right)}+\frac{z}{z+\frac{3}{2}\left(x+y\right)}\)
\(=\frac{2x}{2x+3\left(y+z\right)}+\frac{2y}{2y+3\left(z+x\right)}+\frac{2z}{2z+3\left(x+y\right)}\)
\(=\frac{2x^2}{2x^2+3xy+3xz}+\frac{2y^2}{2y^2+3yz+3xy}+\frac{2z^2}{2z^2+3zx+3yz}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+6\left(xy+yz+zx\right)}=\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+2\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+\frac{2}{3}\left(x+y+z\right)^2}=\frac{2\left(x+y+z\right)^2}{\frac{8}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)
\(\Rightarrow1-3A\ge\frac{3}{4}\Rightarrow A\le\frac{3}{4}\left(Q.E.D\right)\)
Cho x,y,z>0. CMR:
\(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)\)
Cho x, y, z dương. Chứng minh rằng: \(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}.\left(x+y+z\right)\)
CMR: \(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)\)
Với x;y;z > 0.
Giải hệ phương trình :
\(\hept{\begin{cases}x+y+z=6\\\frac{xy+yz+zx}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=2\sqrt{2}\end{cases}}\)
\(\sqrt{x^2+xy+y^2}=\sqrt{\left(x+y\right)^2-xy}\ge\sqrt{\left(x+y\right)^2-\frac{1}{4}\left(x+y\right)^2}=\frac{x+y}{2}.\sqrt{3}\)
cmtt=>\(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)=3\)
