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

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!

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 $$\frac{x^{10}}{32}<\left(1+\frac{x^2}{2}\right)^5,$$ $$x^{10}<32\left(1+\frac{x^2}{2}\right)^5,$$ $$\frac{x^{10}}{\left(1+\frac{x^2}{2}\right)^5}<32,$$ $$\frac{|x^5|}{\sqrt{\left(1+\frac{x^2}{2}\right)^5}}<4\sqrt{2}.$$