Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm having trouble with an exercise in the Cauchy Schwarz Master Class by Steele. Exercise 1.3b asks to prove or disprove this generalization of the Cauchy-Schwarz inequality:

enter image description here

The following is the solution at the end of the book:

enter image description here

After struggling to understand the solution for a few hours, I still cannot see why the substitution $c_k^2 / (c_1^2 + \ldots + c_n^2)$ would bring the target inequality to the solvable inequality. Neither do I understand what the $n^2 < n^3$ bound has to do with anything or how it allows us to take a "cheap shot".


Edit: I'm also wondering, is there a name for this generalization of Cauchy-Schwarz? Any known results in this direction?

share|cite|improve this question
up vote 4 down vote accepted

I have reason to believe the text has a typo; maybe someone can correct me on this point. Because, to my mind, the definition of the $\hat{c}_i$'s would apply to the sum $\sum |a_k b_k c_k^2|$. I suspect it should read


If my hunch is correct, we would argue as follows (using the $\hat{c}_i$'s defined right above):

$$\left|\sum_{k=1}^n a_kb_k\hat{c}_k \right| \color{Red}\le \sum_{k=1}^n |a_kb_k \hat{c}_k| \color{Green}\le \sum_{k=1}^n |a_kb_k| \color{Blue}=\left|\sum_{k=1}^n |a_k|\cdot|b_k|\right| \color{Purple}\le \left(\sum_{k=1}^n |a_k|^2\right)^{1/2}\left(\sum_{k=1}^n |b_k|^2\right)^{1/2} $$


  • $\color{Red}\le$: Follows from triangle inequality
  • $\color{Green}\le$: Follows from $|\hat{c}_k|\le1$, $k=1,\cdots,n$.
  • $\color{Blue}=$: Follows because $x=|x|$ for $x\ge0$.
  • $\color{Purple}\le$: Cauchy-Schwarz applied to $|a_k|,|b_k|$.

Now take the far left and far right side of this, square, and multiply by $c_1^2+c_2^2\cdots+c_n^2$ (apply to $\hat{c}_i$).

share|cite|improve this answer
You are right about the typo. It is listed on the errata page maintained by the author:… – Byron Schmuland Feb 20 '12 at 15:56
Ah, that makes much more sense. It didn't even occur to me that there might be a typo -- I should check errata next time! – Lucky Feb 20 '12 at 16:21

It seems to me that it's much easier to ignore the hint and note that $\left( \sum_{k=1}^{n}(a_k b_k) c_k \right)^{2} \leq \left( \sum_{k=1}^{n} (a_{k} b_{k})^{2} \right) \left( \sum_{k=1}^{n} c_{k}^{2} \right)$ by the usual Cauchy-Schwarz inequality, while it is clear that $\sum_{k=1}^{n} (a_{k} b_{k})^{2} \leq \left(\sum_{k=1}^{n} a_{k}^{2} \right) \left(\sum_{k=1}^{n} b_{k}^{2}\right).$

share|cite|improve this answer

In general, $\sum |s_i t_i| \le \sum |s_i| \sum |t_i|$.

Here, $\sum |\hat c_i| = 1$, since you've essentially defined

$$\hat c_i = \frac{c_i^2}{\sum c_i^2}$$

So in the above, set $s_i = a_ib_i,\ t_i = \hat c_i$. Then apply Cauchy-Schwarz to what remains.

share|cite|improve this answer

When solving the problem, probably one of the first things that you should try to do is is to reduce the problem to the "ordinary" C-S inequality.

The proof above has an aesthetic appeal, but maybe an clearer approach would be to try the following. (At least for me it was)

What would be one of the first things to do if you wanted to reduce the problem to the ordinary C-S ineq.?

Maybe to define $d_n=b_nc_n$ and apply the inequality to the sequences $a_n$ and $d_n$.

What do you get?

Now use that $b_1^2c_1^2+…+b_n^2c_n^2$ is less than $(b_1^2+…+b_n^2)(c_1^2+…+c_n^2)$ since everything is squared.

If you want me to elaborate on the above ideas, please let me know.

share|cite|improve this answer

In my mind, the author want to given an general point of view that in the proof of inequalities, the stronger the inequalities are, the harder to prove it, and since the inequality given in problem is not so strong, we may have many way to reach a proof, and in fact the author has showed that, in order to prove the given inequality, we only need to observe the following tow facts:

  1. For any $a_k, b_k, c_k$ $$ \sum_{k=1}^n a_k b_k c_k\leq \sum_{k=1}^n |a_k| |b_k| |c_k|, $$ thus, we can suppose that $a_k,b_k,c_k>0$ (or replace them with their $|\cdot|$);
  2. Now, the given inequality can be obtained if we have $$ \sum_{k=1}^na_k b_kc_k\leq\left(\sum_{k}a^2_k\sum_k b^2_k\sum_k c^2_k\right)^{1/2} =\left(\sum_{k}a^2_k\sum_k b^2_k\right)^{1/2}\left(\sum_k c^2_k\right)^{1/2}, $$ i.e., $$ \sum_{k=1}^na_k b_k\tilde c_k\leq\left(\sum_{k}a^2_k\sum_k b^2_k\right)^{1/2}, $$ where $$ \tilde c_k=\frac{c_k}{\left(\sum_k c^2_k\right)^{1/2}}. $$ Or just as the first answer shows that $$ \tilde c_k^2=\frac{c_k^2}{\sum_k c^2_k}, $$ this is where you get stuck.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.