An inequality involving $a_i, b_i$ such that $\sum_{i=1}^n a_i = \sum_{i=1}^n b_i = 1$

Question

I am having the hardest time proving this inequality:

Let $n$ be a natural number, and $a_1,a_2,\cdots,a_n$ and $b_1, b_2, ..., b_n$ be nonnegative numbers such that $\sum_{i=1}^n a_i = \sum_{i=1}^n b_i = 1$. Then $$\sum_{i=1}^n \sqrt{a_i b_i} \le 1 - \frac{\big(\sum_{i=1}^n |a_i - b_i|\big)^2}{8}.$$

The proof must be related to Cauchy-Schwarz inequality, but I seem not to be able to connect the dots. Any idea?

Yes, but it did not lead me to a useful method for the general case. — VividD, Jul 10 '17 at 20:35
BTW, this inequality and its continuous counterpart have some applications in classification problems of machine learning. — VividD, Jul 10 '17 at 20:37

score 3 · Accepted Answer · answered Jul 11 '17 at 04:11

3

Let me write $a_j=c_j^2$, $b_j=d_j^2$. Then $$ \sum |c_j^2-d_j^2|=\sum (c_j+d_j)|c_j-d_j| \le \| c+d\|_2\|c-d\|_2\le 2\|c-d\|_2 , $$ by CS and since $c,d$ are unit vectors in $\ell^2$. So the square of the sum on the RHS of your inequality is estimated by $4\|c-d\|_2^2=8(1-\langle c,d\rangle)$, as desired.

answered Jul 11 '17 at 04:11

You probably meant "... square of the sum on the LHS* ..."* – Martin R Jul 11 '17 at 07:58
@MartinR: No, I do mean RHS ("of your inequality" = of the inequality in the OP). – Jul 11 '17 at 13:02
Thanks, to complete the answer, can you perhaps summarize the conditions when equality holds? – VividD Jul 11 '17 at 16:59
When $c=d$ because we can't get equality in the last $\le$ otherwise. – Jul 12 '17 at 18:23

An inequality involving $a_i, b_i$ such that $\sum_{i=1}^n a_i = \sum_{i=1}^n b_i = 1$

1 Answers1