[Math] Help understanding proof of generalization of Cauchy-Schwarz Inequality

inequalityrecreational-mathematics

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".

Thanks!

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

Best Answer

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

$$\hat{c}_i=\frac{c_i}{\sqrt{c_1^2+c_2^2+\cdots+c_n^2}}.$$

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

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

Above:

  • $\color{Red}\le$: 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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