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
The best result I could get :


  $\left(1-\sum_{i=1}^{n}a^3_{i}\right)^{\frac{1}{3}}\cdot
\left(1-\sum_{i=1}^{n}b^3_{i}\right)^{\frac{1}{3}}\ge 
1-\sum_{i=1}^{n}a_{i}b_{i}-\sum_{i=1}^{n}|a_{i}-b_{i}| - \sum_{i=1}^{n} \sqrt{2 a_{i}b_{i}}$


The method I used in trying to prove the claim: 

Let's take $0\leq x\leq1 $ we have $0\leq 1-x\leq1 $ and : $$(1-x)+x \leq 1 $$

If we take the both parts to powers $ m \geq 1 $ they can only get smaller so :
$$ \forall\ m \geq 1 , (1-x)^{m}+x^{m} \leq 1 $$

$$ \forall\ m \geq 1 , (1-x)^{m} \leq 1 - x^{m}$$

lets put $x = \sum x_{i}$ where $x_{i}$ are positive numbers :

$$ \forall\ m \geq 1 ,(1-\sum x_{i})^{m} \leq 1 - (\sum x_{i})^{m} \leq 1 - \sum x_{i}^{m}$$


  We have then our lemma : 
  
  $$ \sum x_{i} \leq 1 \Rightarrow  \forall\ m \geq 1 , \big(1 - \sum x_{i}^{m}\big)  \geq \big(1-\sum x_{i}\big)^m $$
  and if $ \big(1 - \sum x_{i}^{m}\big) \ge 0 $ and $\big(1 - \sum x_{i}\big) \leq 0 $ we still have :
  $$ \forall\ m \geq 1 , \big(1 - \sum x_{i}^{m}\big)^{1/m}  \geq \big(1-\sum x_{i}\big)$$


Back to our problem :
 $$ L = \left(1-\sum_{i=1}^{n}a^3_{i}\right)^{\frac{1}{3}}\cdot
\left(1-\sum_{i=1}^{n}b^3_{i}\right)^{\frac{1}{3}} = \left(1-\sum_{i=1}^{n} (a^{2}_{i})^{\frac{3}{2}}\right)^{\frac{1}{3}} \cdot
\left(1-\sum_{i=1}^{n} (b^{2}_{i})^{\frac{3}{2}}\right)^{\frac{1}{3}}$$

$$ \ge \left(1-\sum_{i=1}^{n} a^{2}_{i} \right)^{\frac{1}{2}} \cdot
\left(1-\sum_{i=1}^{n} b^{2}_{i} \right)^{\frac{1}{2}} $$

(We got here by lemma for $m = 3/2$ and $x_i = a_i^2$)

$$ \ge \left(1 + \sum_{i=1}^{n} a^{2}_{i} b^{2}_{i} + \sum_{i \not=j}^{n} a^{2}_{i} b^{2}_{j}-\sum_{i=1}^{n} a^{2}_{i} -\sum_{i=1}^{n} b^{2}_{i} \right)^{\frac{1}{2}} $$

$$  = \left(1 - \sum_{i=1}^{n} a^{2}_{i} b^{2}_{i} + 2\sum_{i=1}^{n} a^{2}_{i} b^{2}_{i} + \sum_{i \not=j}^{n} a^{2}_{i}b^{2}_{j} -\sum_{i=1}^{n} ( a^{2}_{i} + b^{2}_{i} ) \right)^{\frac{1}{2}} $$

$$  = \left(1 - \sum_{i=1}^{n} a^{2}_{i} b^{2}_{i}  -\sum_{i=1}^{n} ( a^{2}_{i} + b^{2}_{i} -2a_{i}b_{i}) + 2\sum_{i=1}^{n} a^{2}_{i} b^{2}_{i} + \sum_{i \not=j}^{n} a^{2}_{i}b^{2}_{j} - 2\sum_{i=1}^{n} a_{i} b_{i} \right)^{\frac{1}{2}} $$

I blocked here, I WAS NOT ABLE to prove that (Wich is false by the way see here) :

$$ 2\sum_{i=1}^{n} a^{2}_{i} b^{2}_{i} + \sum_{i \not=j}^{n} a^{2}_{i}b^{2}_{j} - 2\sum_{i=1}^{n} a_{i} b_{i} \geq 0 $$

But we can easily prove that (using the lemma): 

$$  L \ge 1 - \sum_{i=1}^{n} a_{i} b_{i} -\sum_{i=1}^{n} ( a^{2}_{i} + b^{2}_{i} - 2 a^{2}_{i} b^{2}_{i})^{\frac{1}{2}} $$

OR :


  $$  L \ge 1 - \sum_{i=1}^{n} a_{i} b_{i} -\sum_{i=1}^{n} | a_{i} + b_{i}|  $$


Or 


  $\left(1-\sum_{i=1}^{n}a^3_{i}\right)^{\frac{1}{3}}\cdot
\left(1-\sum_{i=1}^{n}b^3_{i}\right)^{\frac{1}{3}}\ge 
1-\sum_{i=1}^{n}a_{i}b_{i}-\sum_{i=1}^{n}|a_{i}-b_{i}| - \sum_{i=1}^{n} \sqrt{2 a_{i}b_{i}}$


Edit : Here is my attempt to fix it :

$$ L \geq \left(1 - \sum_{i=1}^{n} a^{2}_{i} b^{2}_{i}  -\sum_{i=1}^{n} ( a^{2}_{i} + b^{2}_{i} -2a_{i}b_{i}) + 2\sum_{i=1}^{n} a^{2}_{i} b^{2}_{i} + \sum_{i \not=j}^{n} a^{2}_{i}b^{2}_{j} - 2\sum_{i=1}^{n} a_{i} b_{i} \right)^{\frac{1}{2}} $$

Let's put :
$$ X = 1 - \sum_{i=1}^{n} a^{2}_{i} b^{2}_{i}  -\sum_{i=1}^{n} ( a^{2}_{i} + b^{2}_{i} -2a_{i}b_{i}) $$

$$ R = 1 - \sum_{i=1}^{n} a_{i} b_{i}  -\sum_{i=1}^{n} | a_{i} - b_{i} | $$
and

$$ Y = 2\sum_{i=1}^{n} a^{2}_{i} b^{2}_{i} + \sum_{i \not=j}^{n} a^{2}_{i}b^{2}_{j} - 2\sum_{i=1}^{n} a_{i} b_{i} $$


if $R &lt; 0 $ the originale inequality holds : OK


if $0 \leq R \leq 1 $ then $0 \leq X \leq 1 $ and we have $ 0 \leq X + Y \leq 1 $ so $ -1 \leq Y \leq 1 $


if $Y &gt; 0$: OK


$(X+Y)^{1/2} \ge \sqrt {X} $ then by the lemma we have the wanted inequality.


if $Y &lt; 0 $ : NOT OK


$(X+Y)^{1/2} = (\sqrt X ^2-\sqrt{-Y}^2)^{1/2} = ((\sqrt X -\sqrt{-Y})(\sqrt X +\sqrt{-Y}))^{1/2} \ge  ? $ 

.. then .. ?? :(
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:

$1-\sum_{i}a_{i}^{3}\geq \sum_{i}a_{i}^{2}-\sum_{i}a_{i}^{3}=\sum_{i}a_{i}(1-a_{i})$