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

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?

• Have you tried the case $n = 2$? – DeepSea Jul 10 '17 at 20:33
• 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
• What can you say about $\sum|\sqrt{a_i}-\sqrt{b_i}|^2$ ? – kimchi lover Jul 10 '17 at 22:45

## 1 Answer

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.

• You probably meant "... square of the sum on the LHS ..." – Martin R Jul 11 '17 at 7:58
• @MartinR: No, I do mean RHS ("of your inequality" = of the inequality in the OP). – user138530 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. – user138530 Jul 12 '17 at 18:23