![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}\left|x\right|\ge3\\\left|y\right|\ge3\\\left|z\right|\ge3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|\dfrac{1}{x}\right|\le\dfrac{1}{3}\\\left|\dfrac{1}{y}\right|\le\dfrac{1}{3}\\\left|\dfrac{1}{z}\right|\le\dfrac{1}{3}\end{matrix}\right.\)
\(\left|A\right|=\left|\dfrac{xy+yz+xz}{xyz}\right|=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\le\left|\dfrac{1}{x}\right|+\left|\dfrac{1}{y}\right|+\left|\dfrac{1}{z}\right|\le\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)
\(\Rightarrow A\le\left|A\right|\le1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dùng BĐT Cosi ta có:
\(\frac{x^3}{yz}+y+z\ge3\sqrt[3]{\frac{x^3}{yz}\cdot y\cdot z}=3x\)
\(\frac{y^3}{xz}+z+x\ge3\sqrt[3]{\frac{z^3}{zx}\cdot z\cdot x}=3y\)
\(\frac{z^3}{yx}+x+y\ge3\sqrt[3]{\frac{z^3}{xy}\cdot x\cdot y}=3z\)
\(\Rightarrow\frac{x^3}{xy}+y+z+\frac{y^3}{zx}+x+z+\frac{z^3}{xy}+x+y\ge3x+3y+3z\)
\(\Rightarrow\frac{x^3}{yz}+\frac{y^3}{xz}+\frac{z^3}{xy}\ge3\left(x+y+z\right)-2\left(x+y+z\right)\)\(=x+y+z\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{x^3}{yz}=y=z\\\frac{y^3}{zx}=x=z\\\frac{z^3}{yz}=y=x\end{cases}\Rightarrow x=y=z}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bđt Cauchy-Schwarz và AM-GM:
\(x^4+y^4+z^4\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3}\ge\dfrac{\left(xy+yz+xz\right)^2}{3}=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta chứng minh được
\(a^2+b^2+c^2\) ≥\(\dfrac{\left(a+b+c\right)^2}{3}\)
A=\(x^4+y^4+z^4\) ≥ \(x^2y^2+y^2z^2+z^2x^2\) ≥ \(\dfrac{\left(xy+yz+zx\right)^2}{3}=\dfrac{1}{3}\)
=>\(x^4+y^4+z^4\) ≥ \(\dfrac{1}{3}\left(đpcm\right)\)
Chúc Bạn Học Tốt
![](https://rs.olm.vn/images/avt/0.png?1311)