Can anyone point me to any reference on the following inequality: Let $c>0, a,b \in \Bbb{R}$ $$ \|a+b\|^2 \leq (1+c)\|a\|^2+(1+{\textstyle\frac{1}{c}})\|b\|^2 $$

I'm not even sure if "Bohr's Inequality" is this inequality's name, since I can't find it anywhere in the internet.

I'm having a bad day and I got stuck in this one, apparently. Any suggestions or hints on how to prove it will be appreciated.

  • $\begingroup$ For a reference (but no proof), see Hardy, Littlewood, and Pólya's Inequalities, second edition, Problem #59 on page 61. They do give Bohr as a reference for the inequality, and they also say that $a$ and $b$ could be real or complex. $\endgroup$ – Mike Spivey Mar 8 '11 at 0:38
  • $\begingroup$ Is there any online repository where I could look that up? It's a holyday in my country, so I can't go look at a library. $\endgroup$ – Leonardo Fontoura Mar 8 '11 at 0:52
  • $\begingroup$ You should be able to access it via Google books: books.google.com. Just search for "Hardy, Littlewood, Polya." $\endgroup$ – Mike Spivey Mar 8 '11 at 2:33
  • $\begingroup$ Thanks, I had actually tried googling that book before to no avail, even thought it is in google's own network. $\endgroup$ – Leonardo Fontoura Mar 8 '11 at 3:05

Here's a hint. $a,b$ and $c$ are real, and the inequality boils down to showing that

\begin{equation} c|a|^2 + \frac{1}{c}|b|^2 -2ab \geq 0 \end{equation}

Consider the function

\begin{equation} f(x) = x|a|^2 + \frac{1}{x}|b|^2 -2ab \end{equation}

What is the minimum value of this function?

  • $\begingroup$ Thanks for the quick answer! I forgot to state that I can't use derivatives here. I can't recall any way of showing the minimum value of a function without using derivatives. $\endgroup$ – Leonardo Fontoura Mar 8 '11 at 0:48
  • $\begingroup$ After thinking a bit over your first hint, you can rearrange it to be $0 \leq \|a\|^2-\frac{2ab}{c} \|b\|^2 = (a-\frac{b}{c})^2$ which is true. Thanks! $\endgroup$ – Leonardo Fontoura Mar 8 '11 at 1:49

Do you mean

$$(a+b)^2 \le (1+c)a^2 + (1 + 1/c)b^2$$ ?

This follows from $\text{AM} \ge \text{GM}$

$$ca^2 + b^2/c \ge 2 ab$$

No add $a^2 + b^2$ to both sides.

If you meant $\mathbb{R}^n$ and the Euclidean Norm, you can prove it by applying the above $n$ times.

  • $\begingroup$ Since a and b are real numbers, $\| \circ \|$ is the absolute value. Thanks for the quick answer. $\endgroup$ – Leonardo Fontoura Mar 8 '11 at 0:50

This is a remark/comment:

The inequality you mention is perhaps be an due to Harald Bohr, there is however an other inequality usually called Bohr's inequality.

Consider a bounded analytic function $f$ on the unit disc, that is $$f(z)=\sum_{n=0}^\infty a_nz^n\ \text{ and }\ \|f\|_\infty=\sup_{|z|<1}|f(z)|<\infty.$$ Then for $0<r\le\frac{1}{3}$ we have (Bohr's inequality) $$\sum_{n=0}^\infty |a_n|r^n\le \|f\|_\infty.$$ Moreover, the constant $\frac{1}{3}$ is the best possible which was actually not proved by Bohr but by F. Wiener.

H. P. Boas and D. Khavinson, Bohr's power series theorem in several variables, Proc. Amer. Math. Soc. 125(1997), 2975-2979.

H. Bohr, A theorem concerning power series, Proc. London Math. Soc. 2(13)1914, 1-5.

  • $\begingroup$ @J.M.: Thanks. $\ \ \ $ $\endgroup$ – AD. Oct 15 '11 at 13:38

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