Bất phương trình thi vào 10

T

thokhoaitay95

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B

braga

Cách khác:
Theo bất đẳng thức [TEX]Caychy-schwarz[/TEX] ta có:
[TEX]\frac{x^4}{y+3z}+\frac{y^4}{z+3x}+\frac{z^4}{x+3y}\geq \frac{(x^2+y^2+z^2)^2}{4(x+y+z)} \\ Do \ 3(x^2+y^2+z^2)\geq (x+y+z)^2 \ \ nen \ \ \frac{(x^2+y^2+z^2)^2}{4(x+y+z)}\geq \frac{(x+y+z)^4}{36(x+y+z)} \\ Ma \ \ (x+y+z)^2\geq 3(xy+yz+zx)=9 \Rightarrow x+y+z\geq 3 [/tex]
[tex]\Rightarrow \frac{(x+y+z)^4}{36(x+y+z)}=\frac{(x+y+z)^3}{36} [/TEX] [TEX]\geq \frac{3^3}{36}=\frac{3}{4} [/TEX]
 
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