How to prove $(1-\sum_{i=1}^{n}a^3_{i})^{1/3}\cdot (1-\sum_{i=1}^{n}b^3_{i})^{1/3}\ge 1-\sum_{i=1}^{n}a_{i}b_{i}-\sum_{i=1}^{n}|a_{i}-b_{i}|$? 
let $a_{1},a_{2},\cdots,a_{n},b_{1},b_{2},\cdots,b_{n}$ be postive numbers, and such
  $$\sum_{i=1}^{n}a^2_{i}\le 1,\sum_{i=1}^{n}b^2_{i}\le 1$$
  show that
  $$\left(1-\sum_{i=1}^{n}a^3_{i}\right)^{\frac{1}{3}}\cdot
\left(1-\sum_{i=1}^{n}b^3_{i}\right)^{\frac{1}{3}}\ge 
1-\sum_{i=1}^{n}a_{i}b_{i}-\sum_{i=1}^{n}|a_{i}-b_{i}|$$

My idea:
since $$a_{i},b_{i}\in (0,1]\Longrightarrow |a_{i}-b_{i}|\le 1\Longrightarrow |a_{i}-b_{i}|\ge |a_{i}-b_{i}|^2=a^2_{i}+b^2_{i}-2a_{i}b_{i}$$
so
$$RHS\le 1-\sum_{i=1}^{n}a_{i}b_{i}-\sum_{i=1}^{n}|a_{i}-b_{i}|^2=1+\sum_{i=1}^{n}a_{i}b_{i}-\sum_{i=1}^{n}a^2_{i}-\sum_{i=1}^{n}b^2_{i}$$
so we only prove
$$\left(1-\sum_{i=1}^{n}a^3_{i}-\sum_{i=1}^{n}b^3_{i}+\sum_{i=1}^{n}a^3_{i}\cdot\sum_{i=1}^{n}b^3_{i}\right)^{1/3}\ge 1+\sum_{i=1}^{n}a_{i}b_{i}-\sum_{i=1}^{n}a^2_{i}-\sum_{i=1}^{n}b^2_{i}$$
$$\Longleftrightarrow 1-\sum_{i=1}^{n}a^3_{i}-\sum_{i=1}^{n}b^3_{i}+\sum_{i=1}^{n}a^3_{i}\cdot\sum_{i=1}^{n}b^3_{i}\ge \left(1+\sum_{i=1}^{n}a_{i}b_{i}-\sum_{i=1}^{n}a^2_{i}-\sum_{i=1}^{n}b^2_{i}\right)^3$$
I think this maybe can use Holder inequality to solve it. And I found this inequality is stronger Holder inequality. Thank you

Hint: because all numbers are positive, the expression is equivalent to:
$(1-\|\vec{a}\|_{3}^{3})^{1/3}
\cdot (1-\|\vec{b}\|_{3}^{3})^{1/3}
\geq 1-\langle \vec{c}+\vec{r};\vec{c}-\vec{r}\rangle
- 2\|\vec{r}\|_{1}$
where $c_{i}=\frac{a_{i}+b_{i}}{2}$ and $r_{i}=\frac{a_{i}-b_{i}}{2}$.
Hint 2:
$1-\sum_{i}a_{i}^{3}\geq \sum_{i}a_{i}^{2}-\sum_{i}a_{i}^{3}=\sum_{i}a_{i}(1-a_{i})$
