Chủ đề này có 4 trả lời

[Rikikudo1102]

cho x,y,z>0 chứng minh 

 

$\frac{x^3}{yz}+\frac{y^3}{xz}+\frac{z^3}{xy}\geq x+y+z$

[kfcchicken98]

giả sử $x\geq y\geq z$

áp dụng bđt trebusep $F\geq \frac{1}{3}(x^{3}+y^{3}+z^{3})(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz})\geq \frac{1}{9}(x^{2}+y^{2}+z^{2})(x+y+z)\frac{9}{xy+yz+xz}\geq \frac{(xy+yz+xz)(x+y+z)}{(xy+yz+xz)}=x+y+z$

[PT42]

$\frac{x^{3}}{yz} + y + z \geq 3. \sqrt[3]{\frac{x^{3}}{yz}. y . z} = 3x$ ......

[Simpson Joe Donald]

cho x,y,z>0 chứng minh 

 

$\frac{x^3}{yz}+\frac{y^3}{xz}+\frac{z^3}{xy}\geq x+y+z$

Theo $Cauchy-Schwarz$ thì:

$\frac{x^3}{yz}+\frac{y^3}{xz}+\frac{z^3}{xy}\ge \frac{(x^2+y^2+z^2)^2}{3xyz}\ge \frac{(xy+yz+zx)^2}{3xyz}\ge \frac{3xyz(x+y+z)}{3xyz}=x+y+z$

[davidsilva98]

Theo bất đẳng thức AM-GM ta có:

$\frac{x^{3}}{yz}+y+z+x\geq 4x$

$\frac{y^{3}}{zx}+z+x+y\geq 4y$

$\frac{z^{3}}{xy}+x+y+z\geq 4z$

=> $\frac{x^{3}}{yz}+\frac{y^{3}}{zx}+\frac{z^{3}}{xy}\geq x+y+z$