Toán 9?

V

vansang02121998

$\dfrac{a^4}{c}+\dfrac{b^4}{d}=\dfrac{(a^2+b^2)^2}{c+d}$

$\leftrightarrow (a^2c-b^2d)^2=0$

$\leftrightarrow a^2d=b^2c \leftrightarrow \dfrac{a^2}{c}=\dfrac{b^2}{d}$

Áp dụng Cauchy

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