Toán 9?

V

vansang02121998

a4c+b4d=(a2+b2)2c+d\dfrac{a^4}{c}+\dfrac{b^4}{d}=\dfrac{(a^2+b^2)^2}{c+d}

↔(a2c−b2d)2=0\leftrightarrow (a^2c-b^2d)^2=0

↔a2d=b2c↔a2c=b2d\leftrightarrow a^2d=b^2c \leftrightarrow \dfrac{a^2}{c}=\dfrac{b^2}{d}

Áp dụng Cauchy

$\dfrac{a^2}{c}+\dfrac{d}{b^2}=b2d+\dfrac{b^2}{d}+\dfrac{d}{b^2} \ge 2$
 
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