$\rightarrow abc \geq 1$
Đặt $a=\dfrac{x}{y}; b=\dfrac{y}{z} \rightarrow c \geq \dfrac{z}{x}$
$\rightarrow \dfrac{1}{\sqrt{a^3+1}}=\sqrt{\dfrac{y^3}{x^3+y^3}}$
Tương tự cộng từng vế
$\sum \dfrac{1}{\sqrt{a^3+1}} \leq \sum \sqrt{\dfrac{y^3}{x^3+y^3}}$
Đặt $x^3=m ; y^3=n ; z^3=p$
$\rightarrow \sum \dfrac{1}{\sqrt{a^3+1}} \leq \sum \sqrt{\dfrac{n}{m+n}}$
$(\sum \sqrt{\dfrac{n}{m+n}})^2$
$=(\sum \sqrt{\dfrac{n(n+p)}{(m+n)(n+p)}})^2$
$\leq (n+p+m+n+m+p)(\sum \dfrac{n}{(m+n)(n+p)})$
$=\dfrac{4(m+n+p)(mn+mp+np)}{(m+n)(n+p)(m+p)}$
$\leq \dfrac{9}{2}$
Vậy ta có dpcm