Show $(\sum | a_i - b_i |)^2 \leq (\sum ( \sqrt{a_i} - \sqrt{b_i} )^2 ) (\sum ( \sqrt{a_i} + \sqrt{b_i} )^2 )$

Question

Let $\{a_i\}^k, \{b_i\}^k$ be nonnegative numbers that sum to $1$. Show that:

$\left (\sum_{i=1}^k \left | a_i - b_i \right | \right) ^2 \leq \left (\sum_{i=1}^k \left ( \sqrt{a_i} - \sqrt{b_i} \right )^2 \right ) \left (\sum_{i=1}^k \left ( \sqrt{a_i} + \sqrt{b_i} \right )^2 \right )$

I have tried applying Jensen's inequality and Cauchy Schwarz, but havent figured out how exactly to apply them. I tried Jensen's by considering the absolute value as $\cdot^{2/2}$ and then using the sum over $a_i$ or $b_i$ as a convex combination.

It's also not clear to me that the condition that each set of numbers sums to $1$ is important in this case (I feel this can be done with just Cauchy Schwarz).

Answer 1

This is a direct application of the CS inequality. Take $s_n=\sqrt{a_n}+\sqrt{b_n}$ and $z_n=\sqrt{a_n}-\sqrt{b_n}$. Then by CS, we have $$\bigg(\sum s_nz_n\bigg)^2\leq \sum s_n^2 \sum z_n^2$$

Answer 2

This is just Cauchy-Schwarz inequality: write $|a-b|$ as $|(\sqrt a-\sqrt b)(\sqrt a+\sqrt b)|$.

The numbers adding up to $1$ is not needed.