Vào làm đi nào!

H

huytrandinh

sử dụng bất đẳng thức cauchy ta có
$\dfrac{a^{4}}{\sqrt[3]{b^{3}+7}}+\dfrac{a^{4}}{\sqrt[3]{b^{3}+7}}+\dfrac{a^{4}}{\sqrt[3]{b^{3}+7}}+\dfrac{b^{3}+7}{16}$
$\geq 2a^{3}$
$=>3\sum \dfrac{a^{4}}{\sqrt[3]{b^{3}+7}}$
$\geq 2\sum a^{3}-\sum \dfrac{a^{3}+7}{16}$
$=\dfrac{31(a^{3}+b^{3}+c^{3})-21}{16}=A$
$.a^{3}+a^{3}+1\geq 3a^{2}$
$=>2(a^{3}+b^{3}+c^{3})\geq 3(a^{2}+b^{2}+c^{2})-3=6$
$<=>a^{3}+b^{3}+c^{3}\geq 3$
$=>A\geq \dfrac{31.3-21}{16}=\dfrac{9}{2}$
$<=>\sum \dfrac{a^{3}}{\sqrt[3]{b^{3}+7}}\geq \dfrac{3}{2}$
$=>min=\dfrac{3}{2}<=>a=b=c=1$
 
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