# How 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

#### Solutions Collecting From Web of "How prove $(1-\sum_{i=1}^{n}a^3_{i})^{1/3}\cdot (1-\sum_{i=1}^{n}b^3_{i})^{1/3}\ge 1-\sum_{i=1}^{n}a_{i}b_{i}-\sum_{i=1}^{n}|a_{i}-b_{i}|$"

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 methode 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 $R < 0$ 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 $Y > 0$: OK

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

• if $Y < 0$ : 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:

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