Bđt trong đề thi HSG 12

H

hthtb22

Áp dụng Holder ta có:
$(x^3+y^3+16z^3)(1+1+\dfrac{1}{4})^2 \ge (x+y+z)^3$
$\Rightarrow P = \dfrac{x^3+y^3+16z^3}{(x+y+z)^3} \ge \dfrac{81}{16}$
Dấu = xảy ra $\Leftrightarrow x=y=z/2$
 
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