Upper bound and limit of $\frac{x^5}{\sqrt{(1+\frac{x^2}{2})^5}}$

Question

Let $Y=\sigma X$ be a scaled Student's t-distributed variabled with scale parameter $\sigma=1/\sqrt{2}$ and $4$ degrees of freedom.

I'm proving that for $k>0$ $$\frac{P(Y>x)}{kx^{-4}}\rightarrow k_0 \qquad\text{for}\ x\rightarrow\infty$$ where $k_0>0$. And to do so I'm using L'Hopital's law for "$0/0$"-limits and the fundamental theorem of calculus. I have that $$f(x)=\frac{\Gamma(5/2)}{\sqrt{2\pi}}\bigg(1+\frac{x^2}{2}\bigg)^{-5/2}$$ is the density function for $Y$. This is where I'm stuck at $$\frac{f(x)}{4kx^{-5}}=\frac{\Gamma(5/2)}{4k\sqrt{2\pi}} \frac{x^5}{\sqrt{\big(1+\frac{x^2}{2}\big)^5}}.$$ How do one prove that $$\frac{x^5}{\sqrt{\big(1+\frac{x^2}{2}\big)^5}}\leq 4\sqrt{2}$$ for all $x\in\mathbb{R}$? This is really all I need, since the lefthand-side defines an increasing function on $\mathbb{R}$, hence this bound is going to be its upper limit. Thanks!

Answer 1

Once we have $$\frac{x^2}{2}<1+\frac{x^2}{2}$$ we can take $5$th power of both sides to get $$\left(\frac{x^2}{2}\right)^5<\left(1+\frac{x^2}{2}\right)^5,$$ then after some rearranging we get the desired result $$\frac{x^{10}}{32}<\left(1+\frac{x^2}{2}\right)^5,$$ $$x^{10}<32\left(1+\frac{x^2}{2}\right)^5,$$ $$\frac{x^{10}}{\left(1+\frac{x^2}{2}\right)^5}<32,$$ $$\frac{|x^5|}{\sqrt{\left(1+\frac{x^2}{2}\right)^5}}<4\sqrt{2}.$$