Toan 9 bat dang thuc

T

transformers123

Cách 1: Theo bđt Schwarz, ta có:

$\dfrac{x^2}{z+y}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y} \ge \dfrac{(x+y+z)^2}{z+y+x+z+x+y} = 1$

Dấu "=" xảy ra khi $x=y=z=\dfrac{2}{3}$

Cách 2: Theo bđt Cauchy, ta có:

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

$\iff \dfrac{x^2}{z+y} \ge x-\dfrac{z+y}{4}$

Làm tương tự rồi cộng lại, ta có:

$\dfrac{x^2}{z+y}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y} \ge (x+y+z)-(\dfrac{x+y}{4}+\dfrac{y+z}{4}+\dfrac{x+z}{4})$

$\dfrac{x^2}{z+y}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y} \ge 1$

Dấu "=" xảy ra khi $x=y=z=\dfrac{2}{3}$
 
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