# How can prove this inequality $|\tau|^{2\gamma} \leq c_5(\gamma) \frac{1+|\tau|}{1+|\tau|^{1-2\gamma}}$?

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.

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.
• (also, as shown in this proof, the result holds for $\gamma < 1/2$, not only $\gamma<1/4$.) – Clement C. Nov 23 '18 at 19:59
• 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. – João Paulo Andrade Nov 23 '18 at 22:38
• @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). – Clement C. Nov 23 '18 at 22:40
• For instance, for $\gamma=1/2$, take $c_5(1/2)=2$. You have $$|\tau| \leq 2\cdot \frac{1+|\tau|}{2}$$ for all $\tau$. – Clement C. Nov 23 '18 at 22:46
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.