Exercise 1.1.5 (Cauchy-Schwarz inequality)

Question

Let $n \geq 1$ and let $a_{1}, \ldots, a_{n}$ and $b_{1}, \ldots, b_{n}$ be real numbers. Verify the identity:

$$
\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2}+\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n}\left(a_{i} b_{j}-a_{j} b_{i}\right)^{2}=\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{j=1}^{n} b_{j}^{2}\right)
$$

and conclude the Cauchy-Schwarz inequality:

$$
\left|\sum_{i=1}^{n} a_{i} b_{i}\right| \leq\left(\sum_{i=1}^{n} a_{i}^{2}\right)^{1 / 2}\left(\sum_{j=1}^{n} b_{j}^{2}\right)^{1 / 2}
$$

Then use the Cauchy-Schwarz inequality to prove the triangle inequality:

$$
\left(\sum_{i=1}^{n}\left(a_{i}^{2}+b_{i}^{2}\right)\right)^{1 / 2} \leq\left(\sum_{i=1}^{n} a_{i}^{2}\right)^{1 / 2}+\left(\sum_{j=1}^{n} b_{j}^{2}\right)^{1 / 2}.
$$

Prakash Anand · Accepted Answer

\begin{proof}
We will prove the first part by induction on $n$.

Base Case $(n=1)$

$$
\left(\sum_{i=1}^{1} a_{i} b_{i}ight)^{2}+\frac{1}{2} \sum_{i=1}^{1} \sum_{j=1}^{n}\left(a_{i} b_{j}-a_{j} b_{i}ight)^{2}=a_{1}^{2} b_{1}^{2}+\frac{1}{2}\left(a_{1} b_{1}-a_{1} b_{1}ight)^{2}=a_{1}^{2} b_{1}^{2}=\left(\sum_{i=1}^{1} a_{i}^{2}ight)\left(\sum_{j=1}^{1} b_{j}^{2}ight)
$$

Thus the base case is true.

Inductive Step. Assume for some $n$ that the ugly mess we want to show is true, is true, and then show it is true for $n+1$

$$
\begin{array}{l}
\left(\sum_{i=1}^{n+1} a_{i} b_{i}ight)^{2}+\frac{1}{2} \sum_{i=1}^{n+1} \sum_{j=1}^{n+1}\left(a_{i} b_{j}-a_{j} b_{i}ight)^{2} \
=\left(\sum_{i=1}^{n}\left(a_{i} b_{i}ight)+a_{n+1} b_{n+1}ight)^{2}+\frac{1}{2}\left[\sum_{i=1}^{n} \sum_{j=1}^{n}\left(a_{i} b_{j}-a_{j} b_{i}ight)^{2}+\sum_{i=1}^{n}\left(a_{i} b_{n+1}-a_{n+1} b_{i}ight)^{2}+\sum_{j=1}^{n}\left(a_{n+1} b_{j}-a_{j} b_{n+1}ight)^{2}ight] \
=\left(\sum_{i=1}^{n} a_{i} b_{i}ight)^{2}+2 \sum_{i=1}^{n}\left(a_{i} b_{i} a_{n+1} b_{n+1}ight)+a_{n+1}^{2} b_{n+1}^{2}+\frac{1}{2}\left(\sum_{i=1}^{n} \sum_{j=1}^{n}\left(a_{i} b_{j}-a_{j} b_{i}ight)^{2}ight)+\sum_{i=1}^{n}\left(a_{i} b_{n+1}-a_{n+1} b_{i}ight)^{2} \
=\left(\sum_{i=1}^{n} a_{i}^{2}ight)\left(\sum_{j=1}^{n} b_{j}^{2}ight)+2 \sum_{i=1}^{n}\left(a_{i} b_{i} a_{n+1} b_{n+1}ight)+a_{n+1}^{2} b_{n+1}^{2}+\sum_{i=1}^{n}\left(a_{i}^{2} b_{n+1}^{2}+a_{n+1}^{2} b_{i}^{2}-2 a_{n+1} b_{i} a_{i} b_{n+1}ight) 	ext { ind. hyp. } \
=\left(\sum_{i=1}^{n} a_{i}^{2}ight)\left(\sum_{j=1}^{n} b_{j}^{2}ight)+2 \sum_{i=1}^{n}\left(a_{i} b_{i} a_{n+1} b_{n+1}ight)+a_{n+1}^{2} b_{n+1}^{2}+\sum_{i=1}^{n}\left(a_{n+1}^{2} b_{i}^{2}+a_{i}^{2} b_{n+1}^{2}ight)-2 \sum_{i=1}^{n}\left(a_{i} b_{i} a_{n+1} b_{n+1}ight)\
=\left(\sum_{i=1}^{n} a_{i}^{2}ight)\left(\sum_{j=1}^{n} b_{j}^{2}ight)+a_{n+1}^{2} b_{n+1}^{2}+\sum_{i=1}^{n}\left(a_{n+1}^{2} b_{i}^{2}+a_{i}^{2} b_{n+1}^{2}ight) \
=\left(\sum_{i=1}^{n} a_{i}^{2}ight)\left(\sum_{j=1}^{n} b_{j}^{2}ight)+\sum_{i=1}^{n}\left(a_{n+1}^{2} b_{i}^{2}ight)+\sum_{i=1}^{n}\left(a_{i}^{2} b_{n+1}^{2}ight)+a_{n+1}^{2} b_{n+1}^{2} \
=\left(\sum_{i=1}^{n} a_{i}^{2}ight)\left(\sum_{j=1}^{n} b_{j}^{2}ight)+a_{n+1}^{2}\left(\sum_{i=1}^{n} b_{i}^{2}ight)+b_{n+1}^{2}\left(\sum_{i=1}^{n} a_{i}^{2}ight)+a_{n+1}^{2} b_{n+1}^{2} \
=\left(\sum_{i=1}^{n} a_{i}^{2}+a_{n+1}^{2}ight)\left(\sum_{i=1}^{n} b_{i}^{2}ight)+b_{n+1}^{2}\left(\sum_{i=1}^{n} a_{i}^{2}+a_{n+1}^{2}ight) \
=\left(\sum_{i=1}^{n+1} a_{i}^{2}ight)\left(\sum_{j=1}^{n+1} b_{j}^{2}ight)
\end{array}
$$

This mess proves the inductive step, so we are done with the first part.

To prove the second part, note that we get the following from the first part (since the other term is greater than or equal to 0 ): $\left(\sum_{i=1}^{n} a_{i} b_{i}ight)^{2} \leq\left(\sum_{i=1}^{n} a_{i}^{2}ight)\left(\sum_{j=1}^{n} b_{j}^{2}ight)$

But now we can take the square root of both sides and we get that:

$\left|\sum_{i=1}^{n} a_{i} b_{i}ight| \leq\left(\sum_{i=1}^{n} a_{i}^{2}ight)^{1 / 2}\left(\sum_{j=1}^{n} b_{j}^{2}ight)^{1 / 2}$

Which was what we wanted

To prove the last part let us do the following:

$$
\begin{aligned}
\sum_{i=1}^{n}\left(a_{i}+b_{i}ight)^{2} &=\sum_{i=1}^{n}\left(a_{i}^{2}+b_{i}^{2}+2 a_{i} b_{i}ight) \
&=\sum_{i=1}^{n}\left(a_{i}^{2}ight)+\sum_{i=1}^{n}\left(b_{i}^{2}ight)+2 \sum_{i=1}^{n}\left(a_{i} b_{i}ight) \
& \leq \sum_{i=1}^{n}\left(a_{i}^{2}ight)+\sum_{i=1}^{n}\left(b_{i}^{2}ight)+2\left(\sum_{i=1}^{n}\left(a_{i}^{2}ight)ight)^{1 / 2}\left(\sum_{j=1}^{n}\left(b_{j}^{2}ight)ight)^{1 / 2} 	ext { From C-S } \
&=\left[\left(\sum_{i=1}^{n}\left(a_{i}^{2}ight)ight)^{1 / 2}+\left(\sum_{j=1}^{n}\left(b_{j}^{2}ight)ight)^{1 / 2}ight]^{2}
\end{aligned}
$$

So we have:

$$
\sum_{i=1}^{n}\left(a_{i}+b_{i}ight)^{2} \leq\left[\left(\sum_{i=1}^{n}\left(a_{i}^{2}ight)ight)^{1 / 2}+\left(\sum_{j=1}^{n}\left(b_{j}^{2}ight)ight)^{1 / 2}ight]^{2}
$$

And now taking the square root of both sides we get:

$$
\left(\sum_{i=1}^{n}\left(a_{i}+b_{i}ight)^{2}ight)^{1 / 2} \leq\left(\sum_{i=1}^{n}\left(a_{i}^{2}ight)ight)^{1 / 2}+\left(\sum_{j=1}^{n}\left(b_{j}^{2}ight)ight)^{1 / 2}
$$

Which was what we wanted.
\end{proof}

Exercise 1.1.5 (Cauchy-Schwarz inequality)

Answers

Comments

Exercise 1.1.5 (Cauchy-Schwarz inequality)

Answers

Comments

Add answer