Prove the inequality... 
Given $2n$ numbers $a_1,a_2,...,a_n;b_1,b_2,...,b_n$, suppose that ${\sum_{j=1}^n a_j} \neq 0$ and ${\sum_{j=1}^n b_j} \neq 0$. Prove that the following inequality :
$({\sum_{j=1}^n {a_j b_j}}) + {\big\{   (\sum_{j=1}^n {a_j^2})(\sum_{j=1}^n {b_j^2})                            \big\} }^{\frac {1} {2} } \geq {\frac {2}{n}} (\sum_{j=1}^n {a_j})(\sum_{j=1}^n {b_j})$
with equality iff
${\frac {a_i} {\sum_{j=1}^n {a_j}}} + {\frac {b_i} {\sum_{j=1}^n {b_j}}} = {\frac{2}{n}} , i=1,2,...,n.$

I realized that  in the "inequality proof" part, I could apply Cauchy-Schwarz inequality on LHS to get :
$({\sum_{j=1}^n {a_j b_j}}) + {\big\{   (\sum_{j=1}^n {a_j^2})(\sum_{j=1}^n {b_j^2})                            \big\} }^{\frac {1} {2} } \geq 2({\sum_{j=1}^n {a_j b_j}})$
How can I prove that $2({\sum_{j=1}^n {a_j b_j}}) \geq {\frac {2}{n}} (\sum_{j=1}^n {a_j})(\sum_{j=1}^n {b_j}) $ ?
 A: 1st step:
First we suppose that $\displaystyle{\sum_{j=1}^{n}a_j^2=\sum_{j=1}^{n}b_j^2=1}$.
In this case we have to prove the inequality: 
$$\displaystyle{\sum_{j=1}^{n}a_jb_j+1\geq\frac{2}{n}\left(\sum_{j=1}^{n}a_j\right)\left(\sum_{j=1}^{n}b_j\right) (1)}$$
We write:
$$\displaystyle{1=\frac{\sum_{j=1}^{n}a_j^2}{2}+\frac{\sum_{j=1}^{n}b_j^2}{2}=\frac{\sum_{j=1}^{n}(a_j^2+b_j^2)}{2}}$$
and therefore it suffices to prove that:
$$\displaystyle{\sum_{j=1}^{n}(a_j+b_j)^2\geq \frac{4}{n}\left(\sum_{j=1}^{n}a_j\right)\left(\sum_{j=1}^{n}b_j\right) (*)}$$
But with the Cauchy-Schwarz inequality for the $n$-tuples $(1,...,1)$, $(a_1+b_1,...,a_n+b_n)$ we get:   
$$\sum_{j=1}^{n}(a_j+b_j)^2\geq\frac{1}{n}\left(\sum_{j=1}^{n}(a_j+b_j)\right)^2=\frac{1}{n}\left(\sum_{j=1}^{n}a_j+\sum_{j=1}^{n}b_j\right)^2$$
Since $(x+y)^2\geq 4xy,\forall x,y \in \mathbb R$ for $\displaystyle{x=\sum_{j=1}^{n}a_j}$ and $\displaystyle{y=\sum_{j=1}^{n}b_j}$ we finally prove $(*)$, as we wanted.
2nd step:
Now, for the general case consider the numbers: $$\displaystyle{x_j=\frac{a_j}{\sqrt{\sum_{j=1}^{n}a_j^2}}},\ \displaystyle{y_j=\frac{b_j}{\sqrt{\sum_{j=1}^{n}b_j^2}},\ j\in \{1,...,n\}}$$ 
Observe that $\displaystyle{\sum_{j=1}^{n}x_j^2=\sum_{j=1}^{n}y_j^2=1}.$ Thus we can apply the already proved inequality $(1)$ for the $x_1,...,x_n,y_1,...,y_n$. Multiply both sides by $\displaystyle{\sqrt{\sum_{j=1}^{n}a_j^2}\sqrt{\sum_{j=1}^{n}b_j^2}}$ and you are done.
For the equalities, you just have to check when equalities occur in the inequalities that I used. It is easy to do it and I leave it to you. 
