Đáp án:
Giải thích các bước giải:
áp dụng cố si ta có
$a^3+$$\dfrac{1}{27}+$$\dfrac{1}{27}≥$$\dfrac{a}{3}$
$b^3+$$\dfrac{1}{27}+$$\dfrac{1}{27}≥$$\dfrac{b}{3}$
$c^3+$$\dfrac{1}{27}+$$\dfrac{1}{27}≥$$\dfrac{c}{3}$
⇒$a^3+$$\dfrac{1}{27}+$$\dfrac{1}{27}+$$b^3+$$\dfrac{1}{27}+$$\dfrac{1}{27}+$$c^3+$$\dfrac{1}{27}+$$\dfrac{1}{27}≥$$\dfrac{a}{3}+$$\dfrac{b}{3}+$$\dfrac{c}{3}$
$a^3+b^3+c^3+$$\dfrac{6}{27}≥$$\dfrac{a+b+c}{3}$
$a^3+b^3+c^3≥$$\dfrac{1}{9}$
dấu bằng xảy ra khi $a=b=c=$$\dfrac{1}{3}$
