Proving Cauchy's inequality: Is this going to work?

I'm reading Steele's The Cauchy-Schwarz Masterclass:

The first problem asks me to prove Cauchy's inequality:

$$a_1b_1+a_2b_2+\cdots+a_nb_n\leq\sqrt{a_1^2+a_2^2+\cdots+a_n^2}\sqrt{b_1^2+b_2^2+\cdots+b_n^2}$$

I've tried to do it by creating a trivial instance of the inequality:

$$ab\leq \sqrt{a^2}\sqrt{b^2}$$

And then thinking about each side as a function:

$$f(a,b)=ab\quad\quad g(a,b)=\sqrt{a^2}\sqrt{b^2}$$

I can see that the image of $f$ is $\mathbb{R}$ while the image of $g$ is $\mathbb{R}_{\geq0}$ then when $a$ or $b$ are positive or negative, I can see that:

$$\begin{matrix} {}&{}&{f}&{g}\\ {-a}&{-b}&{ab}&{ab}\\ {-a}&{+b}&{-ab}&{ab}\\ {+a}&{-b}&{-ab}&{ab}\\ {+a}&{+b}&{ab}&{ab} \end{matrix}$$

And the case when $a\vee b=0$, for which $f=g=0$. Is it enough to prove this inequality? Am I missing something?

I've added more stuff, I have proved the case:

$$a_1b_1+a_2b_2\leq\sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}$$

With the same reasoning, by making a function with both sides:

$$f(a_1,a_2,b_1,b_2)=a_1 b_1 + a_2b_2 \quad \quad g(a_1, a_2, b_1, b_2)=\sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}$$

And pointing that the domain of $f$ is $\mathbb{R}$ and the domain of $g$ is $\mathbb{R_{\geq 0}}$, from here I can deduce that the behavior will be the same for the functions:

$$f(a_1,a_2,b_1,b_2,\cdots , \cdots) \quad \quad g(a_1,a_2,b_1,b_2,\cdots, \cdots)$$

Thus giving values to the function, we can only have:

$$f=g \quad f < g$$

• I do not understand what you are trying to do. How does this show the general case? – Calvin Lin Dec 7 '13 at 16:02
• @CalvinLin Yes. I was also thinking about the same, I guess I just got an answer. I'll have lunch and I'll write when I get back. – Billy Rubina Dec 7 '13 at 16:04
• Schwarz and Buniakovski are going to get mad: that inequality is a particular case of the Cauchy-Schwarz-Bunyakovski inequality. – DonAntonio Dec 7 '13 at 16:29
• @CalvinLin Here it is. – Billy Rubina Dec 10 '13 at 11:23

Consider it to be like that:

$a=(a_1,a_2,...,a_n)$ and $b=(b_1,b_2,...,b_n)$. Then (real numbers)$<a,b>=|<a,b>|\leq \sqrt {<a,a>}\cdot \sqrt {<b,b>}$.

The proof of this is well known.

Hint:If $b=0$ then the proof is trivial.So suppose that $b\neq 0$.

Let $y=\frac {b}{\sqrt {<b,b>}}$. Then compute $<a-λy,a-λy>$ for a $λ\in \Bbb R$. Then by letting $λ=<a,y>$ see what you get from the computation of $<a-λy,a-λy>$.

• I do not think the result you have written is much different from the question... I am confused.. – user87543 Dec 7 '13 at 16:21
• @PraphullaKoushik,i agree. It's the the same.But the difference is that with inner product stuff you can do some tricks. – Haha Dec 7 '13 at 16:46

There are many ways of proving this. One way...

As the inequality is homogeneous in $a_i$, WLOG we can set $\sum a_i^2 = 1$. Similarly, we can set $\sum b_i^2 = 1$. Now we only have to prove:

$$\sum a_i b_i \le 1 \tag{1}$$

From $(a_i-b_i)^2 \ge 0$, we have $2a_i b_i \le a_i^2 +b_i^2$. Adding all of these, we get $(1)$.