Cauchy-Schwarz Inequality

Question

I have to show the following is equivalent to the Cauchy-Schwarz Inequality:

$$\left( \sum (r_i +s_i )^2 \right)^{\frac {1}{2}} \leq \left( \sum r_i ^2 \right)^{\frac {1}{2}} + \left( \sum s_i ^2 \right)^{\frac {1}{2}}$$

The Cauchy-Schwarz Inequality is defined as : $$ \left( \sum {(r_i s_i)} \right)^2 \leq \left( \sum{r_i}^2 \right)\left( \sum{s_i}^2 \right) $$

My Attempt:

Since both sides are non-negative we can square both sides.

$$ \sum (r_i +s_i )^2 \leq \sum {r_i ^2} +2 \left( \sum r_i ^2 \right)^{\frac {1}{2}}\left( \sum s_i ^2 \right)^{\frac {1}{2}} + \sum s_i ^2 $$

Expanding the left-hand side.

$$ \sum (r_i^2 + 2r_is_i + s_i^2 )\leq \sum {r_i ^2} +2 \left( \sum r_i ^2 \right)^{\frac {1}{2}}\left( \sum s_i ^2 \right)^{\frac {1}{2}} + \sum s_i ^2 $$

Using identities of Summation.

$$ \sum r_i^2 + 2\sum {(r_i s_i)} + \sum s_i^2 \leq \sum {r_i ^2} +2 \left( \sum r_i ^2 \right)^{\frac {1}{2}}\left( \sum s_i ^2 \right)^{\frac {1}{2}} + \sum s_i ^2 $$

Cancelling like terms $$ \require{cancel} \cancel {\sum r_i^2} + 2\sum {(r_i s_i)} + \cancel{\sum s_i^2} \leq \cancel{\sum {r_i ^2}} +2 \left( \sum r_i ^2 \right)^{\frac {1}{2}}\left( \sum s_i ^2 \right)^{\frac {1}{2}} + \cancel{\sum s_i ^2} $$

We obtain the following:

$$ 2\sum {(r_i s_i)} \leq 2 \left( \sum r_i ^2 \right)^{\frac {1}{2}}\left( \sum s_i ^2 \right)^{\frac {1}{2}} $$

Dividing by 2: $$ \sum {(r_i s_i)} \leq \left( \sum r_i ^2 \right)^{\frac {1}{2}}\left( \sum s_i ^2 \right)^{\frac {1}{2}} $$

It is at this point that I am stuck. Does squaring both sides preserve the inequality at this point?

Thanks in advance.

Answer 1

Does squaring both sides preserve the inequality at this point?

In a word, "yes": $$ \sum(r_{i} s_{i}) \leq \left|\sum(r_{i} s_{i})\right| \leq \left(\sum r_{i}^{2}\right)^{1/2} \left(\sum s_{i}^{2}\right)^{1/2}, $$ with the second inequality being the square root of "your" Cauchy-Schwarz inequality.

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As a stylistic point, two-column proofs are a (less-than-optimal) habit from school courses that should generally be outgrown if you aspire to be a mathematician. In published work, mathematicians instead build chains of equalities and inequalities. Here, for example, you might argue that "your" C-S inequality implies the "new" one like this: \begin{align*} \sum (r_{i} + s_{i})^{2} &= \sum r_{i}^{2} + 2 \sum (r_{i}s_{i}) + \sum s_{i}^{2} \\ &\leq \sum r_{i}^{2} + 2 \left|\sum (r_{i}s_{i})\right| + \sum s_{i}^{2} \\ &\leq \sum r_{i}^{2} + 2 \left(\sum r_{i}^{2}\right)^{1/2} \left(\sum s_{i}^{2}\right)^{1/2} + \sum s_{i}^{2} \\ &= \left[\left(\sum r_{i}^{2}\right)^{1/2} + \left(\sum s_{i}^{2}\right)^{1/2}\right]^{2}. \end{align*}