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:

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

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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?

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5 Answers5


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 $\hat{c}_i$).

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    You are right about the typo. It is listed on the errata page maintained by the author: http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_errat_Index.html –  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
  • Nice use of colors! – Qi Zhu Mar 13 '19 at 15:51

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 $\sum_{k=1}^{n} (a_{k} b_{k})^{2} \leq \left(\sum_{k=1}^{n} a_{k}^{2} \right) \left(\sum_{k=1}^{n} b_{k}^{2}\right).$

Geoff Robinson
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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.
van abel
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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.

Clive Newstead
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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.

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