How can one prove the inequality $(|r_1s_1|+\cdots+|r_ns_n|)^2\leq(r_1^2+\cdots+r_n^2)(s_1^2+\cdots+s_n^2)$ in $\mathbb{R}^n$?

Question

In the inner product space $\mathbb{R}^n$, Cauchy's inequality tells us that $$ (r_1s_1+\cdots+r_ns_n)^2\leq(r_1^2+\cdots+r_n^2)(s_1^2+\cdots+s_n^2). $$ Apparently the inequality can be improved to $$(|r_1s_1|+\cdots+|r_ns_n|)^2\leq(r_1^2+\cdots+r_n^2)(s_1^2+\cdots+s_n^2). $$ How can this be proven? I attempted with induction on $n$. The base case is clear, and I observed $$ (|r_1s_1|+\cdots+|r_ns_n|+|r_{n+1}s_{n+1}|)^2 $$ can be expanded as $$(|r_1s_1|+\cdots+|r_ns_n|)^2+2(|r_1s_1|+\cdots+|r_ns_n|)|r_{n+1}s_{n+1}|+|r_{n+1}s_{n+1}|^2. $$ On the other hand, $(r+_1^2+\cdots+r_{n+1}^2)(s_1^2+\cdots+s_{n+1}^2)$ can be expanded as $$ (r_1^2+\cdots+r_n^2)(s_1^2+\cdots+s_n^2)+(r_1^2+\cdots+r_{n+1}^2)(s_{n+1}^2)+(s_1^2+\cdots+s_{n+1}^2)(r_{n+1}^2)+r_{n+1}^2s_{n+1}^2. $$ By induction, it then suffices to show $$ 2(|r_1s_1|+\cdots+|r_ns_n|)|r_{n+1}s_{n+1}|\leq(r_1^2+\cdots+r_{n+1}^2)(s_{n+1}^2)+(s_1^2+\cdots+s_{n+1}^2)(r_{n+1}^2). $$ However, I'm not sure this inequality is even true, as even the base case is not clear.

Answer 1

Just note that if $r_i s_i$ is negative for any $i$, you can replace $r_i$ with $-r_i$, and Cauchy's inequality still applies. But the right hand side does not change!

Answer 2

Just replace in the Cauchy inequality $r_i$ and $s_i$ by $|r_i|$ and $|s_i|$ respectively and note that $$|r_i|^2=r_i^2\quad\text{and}\quad |s_i|^2=s_i^2 $$

Answer 3

If Cauchy's inequality is true, it is also true for $t_k = \operatorname{sgn} (r_k s_k) r_k$. Since $t_k s_k = |r_k s_k|$, and $t_k^2 = r_k^2$, the result follows from the original inequality applies to $t_k, s_k$.

Answer 4

First the base case is clear:

$$2|r_1 s_1| |r_2 s_2| \leq r_1^2 s_2^2 + s_1^2 r_2^2$$

because you can subtract the left hand side from both sides, then the inequality becomes $$ |r_1 s_2 - s_1 r_2|^2 \geq 0$$ which is clear.

Next, we start with your inductive assumption. Working with your left hand side: $$ 2(|r_1s_1|+\cdots+|r_ns_n| + |r_{n+1} s_{n+1}|)|r_{n+2}s_{n+2}| = 2(|r_1s_1|+\cdots+|r_{n-1}s_{n-1}|+|r_{n+1}s_{n+1}|)|r_{n+2}s_{n+2}| + 2 |r_nr_{n+2}s_ns_{n+2}|$$ $$\leq(r_1^2+\cdots+r_{n-1}^2+r_{n+1}^2+r_{n+2}^2)(s_{n+2}^2)+(s_1^2+\cdots+s_{n-1}^2+s_{n+1}^2+s_{n+2}^2)(r_{n+2}^2) + 2 |r_nr_{n+2}s_ns_{n+2}|. $$ The last term can be rewritten using the base case: $$2 |r_nr_{n+2}s_ns_{n+2}| \leq r_n^2s_{n+2}^2 + r_{n+2}^2 s_n^2$$ which fills in the missing term from above: $$ 2(|r_1s_1|+\cdots+|r_ns_n| + |r_{n+1} s_{n+1}|)|r_{n+2}s_{n+2}|\leq(r_1^2+\cdots+r_n^2+ r_{n+1}^2 + r_{n+2}^2)(s_{n+2}^2)+(s_1^2+\cdots+s_n^2+s_{n+1}^2 + s_{n+2}^2)(r_{n+2}^2). $$ thus completing the proof by induction.

QED