# Cauchy-Schwarz Inequality

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?

• Perhaps you should write what you call Cauchy Schwarz inequality to. Feb 29, 2016 at 14:17
• The Cauchy-Schwarz Inequality is $\left( \sum (r_i s_i) \right)^2 \leq \left( \sum r_i^2 \right) \left( \sum s_i^2 \right)$
– user316745
Feb 29, 2016 at 14:19
• But then you already have it! Just square both sides in your last line and it is exactly what you cal CS inequality. And yes: squaring preseves the inequality. The other way around can be false, though. Feb 29, 2016 at 14:38
• If that is negative then the inequality is trivially true, and otherwise you can square. The other way around is not generally true, as $\;a^2=b^2\implies a=\pm b\;$ , so equality or inequality cannot be assured. Feb 29, 2016 at 16:19
• Just check it: $\;2^2=4=(-2)^2\;$ , yet $\;2\neq-2\;$, or $\;(-3)^2>2^2\;$ , but $\;(-3)\rlap{\;\,/}>2\;$ Mar 1, 2016 at 13:57

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.

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*}

• By "two-column proof" I mean an argument that starts from one (in)equality and proceeds (as in your question) by performing operations to both sides, arriving at a conclusion. (Or worse, that starts with the desired conclusion and arrives at a tautology. Many "proofs" that $0 = 1$ have this form.) In published papers and most textbooks, you'll find (in)equalities are proven by starting on one side and building a succession of equalities (or "$\leq$"s) leading to the other side. [...] Mar 1, 2016 at 17:07
• I didn't mean to criticize your existing argument, incidentally, but am pointing out that (i) chains of (in)equalities are generally preferable, and (ii) in this case can be made fairly short (e.g., four steps). :) Mar 1, 2016 at 17:08
• I don't mind criticism. I would love to "hear" how one prove that two inequality without the use of the "two-column proof". Would it be simply dealing with the LHS and RHS one at a time?
– user316745
Mar 1, 2016 at 17:17
• The four-step inequality in my answer is (omitting square roots) a proof of one direction. :) Mar 1, 2016 at 22:51