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Does this inequality hold?

Let $n \in \mathbb{N}$. Is it true that $$ \left\lvert \frac{\lVert x \rVert}{\sqrt n \sqrt 3}-\frac{1}{3} \right\rvert \leq \displaystyle\left\lvert \frac{\lVert x \rVert ^2}{n}-\frac{1}{3} \right\rvert .$$

Here $\lVert \cdot \rVert$ denotes 2-norm in $\mathbb R ^n$ and $\lvert \cdot \rvert$ the absolute value in $\mathbb R$.

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  • $\begingroup$ Hint: the RHS is the LHS times $\sqrt{3}(u+1/\sqrt{3})$ where $u+\|x\|/\sqrt{n}$. $\endgroup$
    – Aphelli
    Mar 9, 2021 at 10:29

1 Answer 1

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Note that $$\left| \frac{||x||^2}{n} - \frac{1}{3}\right| = \left| \frac{||x||}{\sqrt n} - \frac{1}{\sqrt 3}\right| \cdot \left| \frac{||x||}{\sqrt n} + \frac{1}{\sqrt 3}\right|$$ while $$\left| \frac{||x||}{\sqrt 3 \sqrt n} - \frac{1}{3}\right| = \frac{1}{\sqrt 3}\left| \frac{||x||}{\sqrt n} - \frac{1}{\sqrt 3}\right| $$ Thus your inequality is equivalent to $$\frac{1}{\sqrt 3} \le \left| \frac{||x||}{\sqrt n} + \frac{1}{\sqrt 3}\right| \ \ \mbox{ or } \ \ \left| \frac{||x||}{\sqrt n} - \frac{1}{\sqrt 3}\right|=0$$ which can be rewritten as $$0 \le ||x|| \ \ \mbox{ or } \ \ ||x|| = \sqrt{\frac{n}{3}}$$ which is clearly true.

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