Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(6\left(x^2+y^2+z^2\right)+10\left(xy+yz+zx\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(=5\left(x+y+z\right)^2+\left(x^2+y^2+z^2\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(\ge5.\left(\frac{3}{4}\right)^2+\frac{\left(x+y+z\right)^2}{3}+\frac{2.9}{4\left(x+y+z\right)}\)
\(=5.\left(\frac{3}{4}\right)^2+\frac{\left(\frac{3}{4}\right)^2}{3}+\frac{2.9}{\frac{4.3}{4}}=9\)
hùi nãy mem nào k sai cho t T_T t buồn
\(VT\ge6\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-2\left(xy+yz+zx\right)+2.\frac{9}{4\left(x+y+z\right)}\)
\(=6\left(x+y+z\right)^2-2.\frac{\left(x+y+z\right)^2}{3}+\frac{9}{2\left(x+y+z\right)}=6.\left(\frac{3}{4}\right)^2-2.\frac{\left(\frac{3}{4}\right)^2}{3}+\frac{9}{2.\frac{3}{4}}\)
\(=\frac{27}{8}-\frac{3}{8}+6=9\)
\(\Rightarrow\)\(VT\ge9\) ( đpcm )
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{4}\)
Chúc bạn học tốt ~
Có \(xy+yz+zx=xyz\)\(\Leftrightarrow\)\(\frac{xy+yz+zx}{xyz}=1\)\(\Leftrightarrow\)\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
\(\frac{x^2y}{y+2x}+\frac{y^2z}{z+2y}+\frac{z^2x}{x+2z}=\frac{1}{\frac{1}{x^2}+\frac{2}{xy}}+\frac{1}{\frac{1}{y^2}+\frac{2}{yz}}+\frac{1}{\frac{1}{z^2}+\frac{2}{zx}}\ge\frac{9}{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}\)
\(=\frac{9}{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}=\frac{9}{1^2}=9\)
Dấu "=" ko xảy ra \(\Rightarrow\)\(\frac{x^2y}{y+2x}+\frac{y^2z}{z+2y}+\frac{z^2x}{x+2z}>9\)
\(\frac{3}{xy+yz+xz}+\frac{3}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{3}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{3}\right)^2}{x^2+y^2+z^2+2xy+yz+xz}=\frac{\left(\sqrt{6}+\sqrt{3}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{3}\right)^2\)
(*) ta CM :\(\left(\sqrt{6}+\sqrt{3}\right)^2>14\)
TA có \(\left(\sqrt{6}+\sqrt{3}\right)^{^2}=6+3+2\sqrt{18}=9+6\sqrt{2}>9+5=14\)
=> \(\frac{3}{xy+yz+xz}+\frac{3}{x^2+y^2+z^2}>14\)
\(VT=6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(=6\left(x+y+z\right)^2-2\left(xy+yz+xz\right)+2\frac{9}{2x+y+z+x+2y+z+x+y+2z}\)
\(\ge6\left(x+y+z\right)^2-2\frac{\left(x+y+z\right)^2}{3}+2\frac{9}{4\left(x+y+z\right)}\)
\(=\: 6\cdot\left(\frac{3}{4}\right)^2-2\cdot\frac{\left(\frac{3}{4}\right)^2}{3}+2\cdot\frac{9}{4\cdot\frac{3}{4}}=9\)