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I'm reading a demonstration that uses this following inequality. For a fixed $\gamma<1/4$, exists a $c_5(\gamma)$, such that

$|\tau|^{2\gamma} \leq c_5(\gamma) \frac{1+|\tau|}{1+|\tau|^{1-2\gamma}}, \forall \tau \in \mathbb{R}$.

I tried to deduce that using the fact that $0<\frac{1}{1+|\tau|}\leq1$, but i couldn't get in anywhere.

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Reorganizing, this is equivalent to proving that the function $f\colon\mathbb{R}\to\mathbb{R}$ given by $$ f(\tau) = \frac{|\tau|+|\tau|^{2\gamma}} {|\tau|+1} $$ is bounded. Note that it is continuous, and $\lim_{+\infty} f = \lim_{-\infty} f = 1$ (using the fact that $2\gamma <1$); therefore, it is bounded.

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  • $\begingroup$ (also, as shown in this proof, the result holds for $\gamma < 1/2$, not only $\gamma<1/4$.) $\endgroup$ – Clement C. Nov 23 '18 at 19:59
  • $\begingroup$ Actually it will not hold for $\gamma < 1/2$, the reason why it holds is because $2\gamma<1-2\gamma$ when $\gamma < 1/4$. I figured this, but it wasn't suffice to prove that. But thank you so much for help me. $\endgroup$ – João Paulo Andrade Nov 23 '18 at 22:38
  • $\begingroup$ @JoãoPauloAndrade Glad it helped, but -- it does hold for all $\gamma \leq 1/2$ (the case $1/2$ is proven similarly, but the limit of $f$ at $\infty$ is $2$, not 1). $\endgroup$ – Clement C. Nov 23 '18 at 22:40
  • $\begingroup$ For instance, for $\gamma=1/2$, take $c_5(1/2)=2$. You have $$|\tau| \leq 2\cdot \frac{1+|\tau|}{2}$$ for all $\tau$. $\endgroup$ – Clement C. Nov 23 '18 at 22:46
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Let $|\tau| = x \geq 0$ and Consider: $$f(x) = \dfrac{x^{2\gamma}+x}{1+x}$$ and prove this has a maximum by taking derivative.

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