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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.

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    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.2011-03-08
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    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.2011-03-08
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    You should be able to access it via Google books: http://books.google.com/. Just search for "Hardy, Littlewood, Polya."2011-03-08
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    Thanks, I had actually tried googling that book before to no avail, even thought it is in google's own network.2011-03-08

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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 $0F. 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.

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    @J.M.: Thanks. $\ \ \ $2011-10-15
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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.

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    Since a and b are real numbers, $\| \circ \|$ is the absolute value. Thanks for the quick answer.2011-03-08
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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?

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    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.2011-03-08
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    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!2011-03-08