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T

trinhminh18

Ta có:
$A^2= \dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}$+
+$\dfrac{2x\sqrt{y}}{\sqrt{z}}+\dfrac{2y\sqrt{z}}{\sqrt{x}}+\dfrac{2z\sqrt{x}}{\sqrt{y}}$
Áp dụng bất đẳng thức Cô-si cho 4 số dương ta có:
$\dfrac{x^2}{y}+\dfrac{x\sqrt{y}}{\sqrt{z}}+\dfrac{x\sqrt{y}}{\sqrt{z}}+z$ > 4x
$\dfrac{y^2}{z}+\dfrac{y\sqrt{z}}{\sqrt{x}}+\dfrac{y\sqrt{z}}{\sqrt{x}}+x$ > 4y
$\dfrac{z^2}{x}+\dfrac{z\sqrt{x}}{\sqrt{y}}+\dfrac{z\sqrt{x}}{\sqrt{y}}+y$ > 4z
\Rightarrow$A^2$> $4(x+y+z)-(x+y+z)=3(x+y+z)$ > 36\Rightarrow A>6
Dấu = khi x=y=z=4
 
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H

huynhbachkhoa23

Theo Cauchy:

$A=\dfrac{2x}{\sqrt{4y}}+ \dfrac{2y}{ \sqrt{4z}}+\dfrac{2z}{\sqrt{4x}} \ge \dfrac{4x}{y+4}+\dfrac{4y}{z+4}+\dfrac{4z}{x+4}$

$=\dfrac{4x^2}{xy+4x}+\dfrac{4y^2}{yz+4y}+\dfrac{4z^2}{zx+4z}$

Theo Cauchy-Schwarz:

$A \ge \dfrac{4(x+y+z)^2}{xy+yz+zx+4(x+y+z)} \ge \dfrac{4(x+y+z)^2}{\dfrac{(x+y+z)^2}{3}+4(x+y+z)}$

$=\dfrac{4}{\dfrac{1}{3}+\dfrac{4}{x+y+z}} \ge \dfrac{4}{\dfrac{1}{3}+\dfrac{4}{12}}=6$

Đẳng thức xảy ra khi $x=y=z=4$
 
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