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}) $ ?

  • $\begingroup$ Why is this tagged as linear-algebra? $\endgroup$
    – user1551
    Jan 13, 2017 at 7:55
  • $\begingroup$ Sorry, I thought linear algebra meant simple high school algebra... $\endgroup$
    – user399078
    Jan 13, 2017 at 7:57
  • $\begingroup$ Then you should use the algebra-precalculus tag. $\endgroup$
    – user1551
    Jan 13, 2017 at 8:00
  • $\begingroup$ Your approach hits a snag - by Chebyshev's inequality, when $a_j, b_j$ are oppositely arranged, we have $$\sum a_j b_j \leqslant \frac1n \left(\sum a_j \right)\left(\sum b_j \right)$$ $\endgroup$
    – Macavity
    Jan 13, 2017 at 11:22
  • $\begingroup$ @Macavity This should NOT happen, provided my approach is "correct"... Is there something wrong with my approach ??? $\endgroup$
    – user399078
    Jan 13, 2017 at 12:26

1 Answer 1


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:


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:


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.

  • $\begingroup$ As the inequality is homogeneous in both $a_i$ and $b_i$, the first step is enough +1. $\endgroup$
    – Macavity
    Jan 16, 2017 at 6:42
  • $\begingroup$ @Macavity You are right. I wanted to give a full explanation for the O.P.. By the way, thanks for the (+1). $\endgroup$ Jan 16, 2017 at 9:29

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